The user is asking me to prove the Wilson prime conjecture. Let me think about what this is and whether I can actually prove it. The Wilson prime conjecture (also known as the Wilson quotient conjecture) concerns Wilson primes. A Wilson prime is a prime $p$ such that $(p-1)! \equiv -1 \pmod{p^2}$, or equivalently, the Wilson quotient $W_p = \frac{(p-1)! + 1}{p} \equiv 0 \pmod{p}$. The known Wilson primes are 5, 13, and 563. The conjecture states that there are infinitely many Wilson primes. This is a famous open problem in number theory. It has NOT been proven. No one has proven that there are infinitely many Wilson primes. This is one of the major open problems related to factorial congruences. So the honest answer is: I cannot prove this conjecture because it is an open problem in mathematics. The conjecture was formulated by... actually, let me recall. The conjecture that there are infinitely many Wilson primes is attributed to various sources. It's related to work by Lerch (1905) who showed that if a certain condition related to the Fermat quotient holds, then $p$ is a Wilson prime. Glaisher (1900) showed the Wilson quotient has many properties similar to the Fermat quotient. It's conjectured that the Wilson quotient behaves "randomly" mod p, so the probability that $W_p \equiv 0 \pmod p$ is about $1/p$, and since $\sum 1/p$ diverges, there should be infinitely many Wilson primes. This heuristic is standard. I should: 1. Explain what the Wilson prime conjecture is 2. Clearly state that this is an open problem and no proof exists 3. Explain the heuristic argument for why it's believed to be true (the $1/p$ probability heuristic, sum of reciprocals of primes diverges) 4. Maybe present what is known: known Wilson primes (5, 13, 563), search bounds (Wilson primes up to very large bounds have been searched — I believe the search has gone up to $2 \times 10^{13}$ or higher without finding more; let me recall. PrimeGrid searched for Wilson primes up to about $5 \times 10^8$ initially, then extended. As of recent work, searches have covered up to $2 \times 10^{13}$? I think the current status is that all Wilson primes below $2^{32}$? Hmm. Let me be careful. I know that searches by PrimeGrid found no new Wilson primes up to about $10^{10}$ or so. I should be cautious with exact figures and hedge appropriately.) Actually, what I recall: The first three Wilson primes are 5, 13, 563. Searches have been conducted to at least $2 \times 10^{13}$ (PrimeGrid). I'm not 100% certain of the exact bound, so I should hedge or just say "extensive searches up to hundreds of trillions" or give a ballpark. Let me also discuss related results: - Lerch's theorem: The number of Wilson primes below $x$ is at least $\log \log \log x$ minus something? Actually, Lerch (1905) proved that the number of primes $p \le x$ with $p$ dividing... hmm, let me recall. Lerch showed that the sum of Wilson quotients is congruent to the number of Wilson primes mod p... Actually the relevant result: Lerch proved that $\sum_{p \le x} W_p$ relates... Let me think. Lerch's result: If $p_k$ is the k-th prime and $W_p$ the Wilson quotient, then the number of Wilson primes less than $p_k$ is congruent to $-\sum_{i=1}^{k-1} W_{p_i} \pmod {p_k}$. This gives that for each fixed number $m$, there are at least $\log\log\log x$ primes $p \le x$... no wait. There's a result (by Lerch, I believe) that the number of Wilson primes up to $x$ is at least of order $\log \log \log x$ infinitely often? Hmm, actually I recall that a consequence of Lerch's congruence is that for any fixed nonnegative integer $m$, the number of Wilson primes up to $x$ exceeds $m$ for $x$ large enough... Something like: the number of Wilson primes $\le x$ tends to infinity? No, that's not proven either. Let me think again. Lerch's congruence: $W_{p_k} \equiv$ ... Hmm. Actually here's the statement I remember: Let $p_k$ be the $k$-th prime. Then the number of Wilson primes among $p_1, \dots, p_{k-1}$ is congruent modulo $p_k$ to $\sum_{i=1}^{k-1} W_{p_i}$ (up to sign). Since $W_{p_i} \ge 1$, this shows that the number of Wilson primes less than $p_k$ is at most... hmm, this gives an upper bound on the number of Wilson primes below $p_k$: it's at most the sum of the Wilson quotients, which isn't immediately useful. Actually, the consequence: the number of Wilson primes less than $p_k$ is at most $\sum_{i 1$, $n$ is prime iff $(n-1)! \equiv -1 \pmod n$. Proof: If $n$ is prime, every nonzero residue mod $n$ has a multiplicative inverse; pair each $a \in \{1,...,p-1\}$ with its inverse $a^{-1}$; the fixed points satisfy $a^2 \equiv 1$, i.e., $a \equiv \pm 1$; so the product $(p-1)! \equiv 1 \cdot (-1) \cdot \prod_{\text{pairs}} (a a^{-1}) \equiv -1 \pmod p$. Conversely, if $n$ composite with a proper divisor $d$, then $d | (n-1)!$, and also... if $(n-1)! \equiv -1 \pmod n$ then $(n-1)!+1 \equiv 0 \pmod n$; since $d | n$, we get $d | (n-1)!+1$ and $d | (n-1)!$ so $d | 1$, contradiction, unless... need $d < n$ so that $d$ appears as a factor in $(n-1)!$: yes since $1 \le d \le n-1$, $d$ is among the factors, so $d | (n-1)!$; then $d | (n-1)!+1 - (n-1)! = 1$, so $d=1$, contradiction. Edge case: $n = 4$: $(3)! = 6 \equiv 2 \pmod 4$, fine, composite case handles $n=4$? $d=2$ divides 6 and $d|n$, then $2 | (6+1)=7$? No, contradiction works. Good. Also mention: Wilson quotient and known data table? Could include the small table: p=5: 4!+1=25, W=5, 5|5 ✓. p=13: 12!+1 = 479001601 = 13^2 × 2834161? Let me verify: 13^2 = 169. 12! = 479001600. +1 = 479001601. 479001601 / 169 = 2834329? Let me compute: 169 × 2,834,000 = 478,946,000. Remainder: 479,001,601 − 478,946,000 = 55,601. 169 × 329 = 55,601. So quotient = 2,834,329. So $W_{13} = 2{,}834{,}329$, and $13 | W_{13}$? 13 × 218,025 = 2,834,325, remainder 4. Hmm that gives remainder 4, but 13 IS a Wilson prime so 13 should divide $W_{13}$. Let me recompute. 12! = 479,001,600. Yes. +1 = 479,001,601. Divide by 13: 479,001,601 / 13 = 36,846,277. Yes since 13 × 36,846,277 = 478, - let me check: 36,846,277 × 13 = 36,846,277 × 10 + ×3 = 368,462,770 + 110,538,831 = 479,001,601. ✓. Then $W_{13} = 36,846,277$. Is it divisible by 13? 36,846,277 / 13 = 2,834,329.0? 13 × 2,834,329 = 36,846,277. ✓ Great. So $W_{13} = 13 × 2,834,329$. I earlier divided by 169 wrongly. Fine. Maybe skip detailed table to avoid arithmetic risk, or present 5 and 13 carefully: $4!+1 = 25 = 5^2$; $12!+1 = 479{,}001{,}601 = 13^2 \cdot 2{,}834{,}329$. Check: $169 \times 2{,}834{,}329 = ?$ 2,834,329 × 169 = 2,834,329 × 170 − 2,834,329 = 481,835,930 − 2,834,329 = 479,001,601. ✓ 563: $563^2 = 316,969$. $562! + 1$ divisible by 316,969 — known fact; I won't verify digits. Heuristic math: expected number up to x: $\sum_{p\le x} 1/p \approx \log\log x + B_1$ (Mertens). At x = 10^13: ln(10^13) ≈ 29.94; ln(29.94) ≈ 3.40. Plus Mertens constant 0.2615 ≈ 3.66. Expected ≈ 3.7 — matches 3 found. To expect 4: need log log x + 0.26 ≈ 4 ⇒ log log x ≈ 3.74 ⇒ log x ≈ e^{3.74} ≈ 42 ⇒ x ≈ e^{42} ≈ 1.7 × 10^{18}. Hmm interesting: including the constant gives ~2×10^{18}. Without constant ~5×10^{23}. So "somewhere between 10^{18} and 10^{24}" — I could say "the heuristic predicts the fourth Wilson prime around 10^{18}–10^{24}, well beyond current searches." Let me double check Mertens constant: $B_1 = 0.2614972...$ yes (Meissel–Mertens constant). $\sum_{p \le x} 1/p = \log\log x + B_1 + o(1)$. At $x = e^{e^4} \approx 5 \times 10^{23}$, sum ≈ 4.26; at $x = e^{42} \approx 2 \times 10^{18}$, sum ≈ 3.74 + 0.26 = 4.0. So "fourth expected by about 10^{18}–10^{23}". Good. Glaisher's congruence precise form: $W_p \equiv B_{p-1} + 1/p - 1 \pmod p$ for $p \ge 5$, where $B_{p-1}$ is the Bernoulli number (rational; its denominator is not divisible by p by von Staudt–Clausen, so $pB_{p-1}$ makes sense mod p... careful: $B_{p-1} + 1/p$ is a rational with denominator $p \cdot \text{den}(B_{p-1})$... The statement: $W_p \equiv B_{p-1} + \frac{1}{p} - 1 \pmod p$, meaning $p | W_p - B_{p-1} - 1/p + 1$ as rationals. Yes that's the standard statement. Consequence: $p$ is Wilson iff $p$ divides numerator of $B_{p-1} + 1/p - 1$. Fine, I can state it without deep dive. Lerch's congruence: with $p_k$ the k-th prime: $\sum_{i=1}^{k-1} W_{p_i} \equiv \#\{i \le k-1 : p_i \text{ Wilson prime}\} \pmod{p_k}$. I think that's right. Let me sanity-check with tiny numbers: $W_2 = (1!+1)/2 = 1$. $W_3 = (2!+1)/3 = 1$. $W_5 = 5$. $W_7 = (720+1)/7 = 721/7 = 103$. $W_{11} = (10!+1)/11 = 3628801/11 = 329891.9? 11 × 329891 = 3,628,801 ✓ so 329,891. Now $p_5 = 11$, $p_6 = 13$. Number of Wilson primes < 13 is 1 (just 5). Sum $W_2+W_3+W_5+W_7 = 1+1+5+103 = 110$. $110 \bmod 13 = 110 - 104 = 6$. But Wilson primes below 13: only 5, count 1. 6 ≠ 1 mod 13. So the naive statement fails! So Lerch's congruence must be something else. Let me recall more carefully. Lerch's theorem (1905): For the $k$-th prime $p_k$, $$\sum_{i=1}^{k-1} \frac{(p_i - 1)! + 1}{p_i} \equiv \#\{\text{Wilson primes } < p_k\} \pmod{p_k}?$$ That's what I wrote and it fails. So maybe the congruence involves $\sum \frac{(p_i-1)!}{p_i}$ without the +1, or maybe it's about $\sum_{i N were Wilson primes. Then for large x, # Wilson primes ~ π(x) ~ x/log x ≫ log log x. The heuristic says ~log log x, but heuristics aren't proofs. Any known theorem bounding # Wilson primes by O(log log x + something)? I strongly doubt it. So the honest claim: "we cannot currently even prove that a positive proportion (or indeed all but finitely many!) primes fail to be Wilson primes" — hmm, "all but finitely many primes are Wilson primes" is consistent with everything we can prove? That seems like a fun striking fact but I should hedge it: I'm not aware of any result ruling it out. Actually hmm, wait: is there something from the Lerch congruence (the real one) that bounds the count? If the real Lerch congruence determines the count mod p_k exactly, it doesn't bound it. OK, hedge: "As far as I know, no unconditional upper bound better than triviality is known — we can't even rule out that almost all primes are Wilson primes." Hmm, is that really true though? Let me think harder. #Wilson primes ≤ x. Suppose p is Wilson. Does that impose constraints on other primes? No mechanism. Trivially # Wilson ≤ π(x). I think indeed no better unconditional upper bound is known. But hmm — actually maybe via the size of $(p-1)!+1$? $(p-1)! + 1$ is divisible by $p^2$; that's it. Product over Wilson primes $p \le x$ of $p^2$ divides $\prod ((p-1)!+1)$ — no constraint linking different p. Right: no upper bound beyond π(x). I'll state it hedged ("to my knowledge, even proving the Wilson primes have density 0 among primes, or that there are infinitely many non-Wilson primes... wait that direction is trivial?" Is it trivial that there are infinitely many NON-Wilson primes? Hmm! If all but finitely many primes were Wilson primes, then non-Wilson primes finite. Can we disprove? Need: infinitely many primes p with $p^2 \nmid (p-1)!+1$. Hmm, is there an easy such family? E.g., primes ≡ ? mod something... The value $(p-1)! \bmod p^2$ — no congruence-class obstruction known. I believe it's genuinely open whether non-Wilson primes are infinite! Wow, that's a nice pedagogical point but I must be confident before asserting. Let me sanity check: is anything known? Search memory: "infinitely many non-Wilson primes" ... I don't recall such a theorem. There's a known theorem: "infinitely many primes p such that (p−1)! ≢ −1 mod p²" — hmm. Honestly can't recall any. The analogous statement for Wieferich: "infinitely many non-Wieferich primes" IS known (unconditionally! — because $2^{p-1} - 1$'s squarefree part... hmm, actually yes: infinitely many primes p with $p^2 \nmid 2^{p-1}−1$: this follows from Silverman? No — Silverman's is conditional. Unconditional: there are infinitely many primes p such that p² ∤ 2^{p−1}−1? Hmm, I recall: it IS known unconditionally that the set of non-Wieferich primes is infinite — is it? If all but finitely many primes were Wieferich, then $2^{p-1} \equiv 1 \pmod{p^2}$ for all large p. Hmm, I don't think we can contradict that... wait, actually I recall a classical result: the number of non-Wieferich primes up to x is at least log x (unconditionally), because... hmm, is that right? There's a result of A. Schinzel? Or is it: $q_p(2)$ distinct mod... For non-Wieferich: suppose $p^2 | 2^{p-1} - 1$ for all p in a set of density 1 — can't be contradicted easily. I recall: "It is not known whether there are infinitely many non-Wieferich primes"? No wait — that can't be: it IS known (I'm fairly sure) that there are infinitely many primes p with $p^2 \nmid 2^{p-1} - 1$. Hmm, why? Consider $2^{p-1} - 1$ for p | something... Think: if $p^2 | 2^{p-1} - 1$, then p is "Wieferich to base 2". Infinitely many non-Wieferich: Suppose all large primes Wieferich. Take the product $M = \prod_{p \le x, p \text{ large}} p^2$ hmm, $2^{\text{lcm stuff}}$... Use order arguments: if $p^2 | 2^{p-1} - 1$ then $p | $ ... hmm. Actually here's a cleaner known statement: the number of primes $p \le x$ for which $p^2 \nmid 2^{p-1} − 1$ is at least $\log x$?? I genuinely recall reading that unconditionally, the count of non-Wieferich primes is ≫ log x, by a clever argument (due to ... maybe using that $\prod_{p \le x}(2^{p-1}-1)$ is not too big?). Hmm, wait: I think the argument is: if $p^2 | 2^{p-1} - 1$ for all primes in some set S ⊆ (y, x]... then consider $N = 2^{\prod_{p \in S}(p-1)} - 1$? Each such p² divides... no: p | 2^{p−1} −1 always; p² | 2^{p−1} − 1 for Wieferich. Consider $2^K - 1$ with K = lcm of (p−1) over p ∈ S: then every p ∈ S divides $2^K − 1$; and if p is Wieferich then p² | 2^{p−1} − 1 | 2^K − 1 (since (p−1) | K). So $2^K − 1$ is divisible by $\prod_{p∈S} p^2$, hence $2^K ≥ \prod p^2$. But $2^K$ is roughly... K = lcm of p−1 which can be huge (e^{x}), so $\prod p^2 ≤ 2^K$ is satisfiable — no contradiction. Hmm so that argument fails; maybe the known result is conditional or I misremember. I recall now: it is a theorem (unconditional) that there are infinitely many primes p such that $p^2 \nmid 2^{p-1} − 1$; moreover ≫ log x of them up to x. I think this is due to ... hmm, "A. S. Bang"? No. Possibly it's in Crandall–Pomerance: "Exercise: show that the number of non-Wieferich primes ≤ x is ≫ log x." Hmm... and there's the stronger: $\gg \log x / \log\log x$? I have a feeling there IS an unconditional lower bound for non-Wieferich primes via an argument using theabc? no, via... Let me try to construct: Suppose $p^2 | 2^{p-1} − 1$. Then the multiplicative order... p² | 2^{p−1} −1 means ord_{p²}(2) | p−1. Fine. Now consider primes q ≡ 1 mod p²... no. Alternative: consider $2^{p−1} ≡ 1 \pmod {p^2}$ ⟺ p² | 2^{p−1} − 1 ⟺ $2^{p−1} ≡ 1 \pmod {p^2}$ ⟺ p is base-2 Wieferich. Known counts: only 2 below 6.7e15. If unconditional lower bound ≫ log x existed for non-Wieferich... yes I'm now quite sure such a result exists! It's an exercise: Let $w(x)$ = # non-Wieferich primes ≤ x. Claim: $w(x) \gg \log x$. Proof sketch (from Crandall–Pomerance or from A. Granville): For each prime p ≤ x, $p$ divides $2^{p-1} − 1$; the Wieferich ones have p² dividing. Consider the product $\prod_{p ≤ x} p$ (primorial) vs $\prod p^2$ for Wieferich... both divide $\prod (2^{p-1} − 1)$ hmm no clean contradiction. Alternatively use $2^{n} − 1$ sequence: Let $N_k = 2^k − 1$. The primes dividing some $N_k$, k ≤ m, are ≫ ... and for each such p, ord_p(2) ≤ m so p ≡ 1 mod ord... For Wieferich relation to squares: p² | 2^{p−1} − 1 implies p² | 2^{ord} ... if ord_p(2) = d then p² | 2^{d·?}... Note $2^{p−1} − 1 = (2^d)^{(p−1)/d} − 1$. If p² | 2^{p−1} − 1 and p ∤ (p−1)/d... by LTE, p² | 2^d − 1 requires... hmm, LTE: $v_p(2^{p-1} − 1) = v_p(2^d − 1) + v_p((p−1)/d)$. Since p ∤ (p−1)/d (as p ∤ p−1), we get p is Wieferich iff p² | 2^d − 1 where d = ord_p(2). So Wieferich ⟺ p² | 2^{ord_p(2)} − 1. Now if p is Wieferich, then p² | 2^{ord} − 1 ⟹ p² ≤ 2^{ord} − 1 ≤ 2^{p−1} − 1... trivial. Hmm, how to get a contradiction from "many Wieferich primes"? Consider q = smallest prime... I think the actual known result: (Silverman 1988, conditional on ABC) # non-Wieferich primes in [1,x] ≫ log x. And UNCONDITIONALLY: hmm, I now recall Crandall–Pomerance state: "It is not even known whether the non-Wieferich primes are infinite"?? No... hmm hmm. Let me think about whether "infinitely many non-Wieferich primes" could be provable: Suppose all primes p ≥ p₀ satisfy $2^{p−1} ≡ 1 \pmod{p^2}$. Is there any known contradiction? ... I don't believe one is known! Wait, but I distinctly remember "it is known that there are infinitely many primes p for which $p^2 \nmid 2^{p−1}−1$" — maybe from the theory of $2^n − 1$ factorizations: if p is Wieferich then p appears in $2^{p−1} − 1$ only to... hmm. Try: Let $A = \prod_{p \le x} (2^{p−1} − 1)$. Then $\prod_{p \le x} p | A$; Wieferich p contribute p². log A ≈ $\sum (p−1)\log 2 ≈ \frac{x^2}{2\log x}\log 2$; log of $\prod p$ ≈ x; so no size contradiction at all — the product is huge. So no easy bound. I'm now doubting the unconditional non-Wieferich infinitude... Actually, I now recall more precisely: it IS known unconditionally that there are infinitely many non-Wieferich primes? Hmm, let me think about the sequence $a_n = 2^n - 1$. Known: the "squarefree part" ... A. Schinzel / others proved: for any integer a>1, infinitely many primes p with $p^2 \nmid a^{p-1} - 1$?? ... I believe YES, this is known, by considering the primes dividing $a^{2^n} - 1$... hmm, why would that help? If p | 2^{2^n} − 1 then ord_p(2) = 2^j, j ≤ n, so p ≡ 1 mod 2^j... p² | 2^{p−1} − 1 ⟺ p² | 2^{2^j} − 1 ⟺ p² | 2^{2^j} − 1 with 2^j ≤ 2^n. So Wieferich primes ≤ x that divide $\prod_{k ≤ n}(2^{2^k} − 1)$ must divide it with multiplicity ≥ 2... The product $\prod_{k≤n} (2^{2^k} −1)$: each prime p ≤ x with ord ≤ 2^n divides it. Hmm, # primes with ord_p(2) ≤ 2^n: these are primes dividing 2^{2^n} − 1... wait ord_p(2) | 2^n iff p | 2^{2^n} − 1. So p ≡ 1 mod ord... The number of prime divisors of 2^{2^n} − 1 (with multiplicity) is ≤ log_2(2^{2^n}) = 2^n. Meanwhile the number of primes p ≤ x with ord_p(2) | 2^n... equals # prime divisors of 2^{2^n} − 1 ≈ could be small. No contradiction either. OK here's the thing — I should stop trying to reconstruct and just be careful: I'll say "closely related questions (infinitude of Wieferich primes, of Mersenne primes, ...) are equally open." For the "we can't even rule out that all but finitely many primes are Wilson primes" claim — hmm, I'd better double check whether something like that is known to be false... I'll soften to: "we cannot currently prove even that the Wilson primes have density zero among the primes." I'm confident THAT is unknown (no technique gives density control here). Hmm, wait: could Lerch's congruence (the real one, whichever it is) imply density bounds? If it determines the count mod p_k exactly, no bound. OK. Also, hmm, actually I want to double-check the statement "even (v)": "no residue class of W_p mod p is provably hit infinitely often." For class 0 = conjecture; class 1 = Lerch primes (3, 103, 839, 2237 known; infinitude open ✓). Class... some classes might be provably empty for small p but generally open. Also, there's a subtlety: W_p mod p for p=2 is 1, p=3 is 1 — but "infinitely often" is the question. I'm confident no class is known infinite. Also for the Fermat quotient: infinitely many p with q_p(2) ≡ a mod p for fixed a — also open (known for... hmm, there are conditional results; "Wieferich primes base 2 with q ≡ a": open). Fine. Now, about search bounds, let me settle on wording: "As of the latest searches I'm aware of (PrimeGrid and predecessors), every prime up to roughly $10^{13}$ has been checked, and no fourth Wilson prime has been found; the exact current bound changes over time." Hmm — is $10^{13}$ right? I recall: McIntosh (2011?) did 5×10^8... hmm wait, actually I think McIntosh's bound was $5 \times 10^8$ around 2004–2011? And PrimeGrid started a Wilson prime search ~2011–2012; by 2018 it passed 10^10; recent reports ~2×10^13 (2023?). There was also a claim "no Wilson primes below 2^32" from an early PrimeGrid milestone (2^32 ≈ 4.3×10^9, reached ~2011?). I'll write "well past $10^{12}$" or "around $10^{13}$" with explicit hedging: "of order $10^{13}$ (check current records — PrimeGrid maintains this search)". I think that's acceptable with the hedge. Now the heuristic section, written carefully: Assume $W_p \bmod p$ is equidistributed in $\{0, 1, \dots, p-1\}$ as p varies (justified by computations & analogy). Then P(p is a Wilson prime) = 1/p. Expected count up to x: $\sum_{p\le x} 1/p = \log\log x + M + o(1)$ (Mertens, M ≈ 0.2615). Since $\sum_p 1/p$ diverges, the model predicts infinitely many. Comparisons: up to 10^13 expected ≈ 3.4 + 0.26 ≈ 3.7 vs 3 observed ✓. Expected 4th by ~e^{e^{3.74}} ≈ 2×10^{18} (using constant) / ~5×10^{23} (rough). So absence of a 4th below 10^13 is unsurprising. Also under this model #Wilson primes ~ log log x: to get 10 of them you'd need x ~ e^{e^{10}} ≈ 10^{9458}!! Fun: $\log\log x = 10$ ⇒ $\log x = e^{10} ≈ 22026$ ⇒ x ≈ e^{22026} ≈ 10^{9566}. So even a 10th Wilson prime would be beyond astronomical. Good illustration of why the conjecture is both plausible and untestable. Also include: The same heuristic predicts infinitely many Wieferich (1093, 3511), Wolstenholme (16843, 2124679), Fibonacci-Wieferich (none known) primes — all open. And Mersenne primes ~ $c\log x$... etc. "Why it's hard" section: - To prove infinitude you'd want some structure forcing $p^2 \mid (p-1)! + 1$ for infinitely many p. The factorial is the "most composite" object; mod p², $(p-1)!$ is governed by a complicated pairing of residues with inverses; the error term mod p² (essentially $W_p$) is expected to behave randomly. There's no known identity that expresses $W_p \bmod p$ through values of L-functions, traces, character sums, or other machinery that analytic number theory can control. Compare: infinitude of primes in AP uses Dirichlet L-functions; infinitude of irregular primes (Carlitz 1954) uses that $\sum_{(a)}$ over Bernoulli numerators has growing span — a "monotone growth" argument; the Wilson condition is a single "diagonal" index (a = p−1) — Carlitz's argument doesn't apply. - Even the complementary statements are open (density zero; hitting any fixed residue class infinitely often). - Concrete known reductions: Glaisher's congruence turns the question into: infinitely many p with $p \mid \mathrm{num}(B_{p-1} + \frac{1}{p} - 1)$ — still no tools. Then close with offers. Also, maybe a brief note: "If you'd like, I can: (1) prove Wilson's theorem rigorously; (2) prove the equivalence $p$ Wilson ⟺ $p | B_{p-1}\cdot$...; (3) rigorously derive the heuristic; (4) sketch searches; (5) review a claimed proof you've seen for where it breaks." And warn: beware of claimed elementary proofs floating around; they all fail (the conjecture implies, e.g., nothing contradictory, but standard techniques provably insufficient... eh, keep short). Now the proof of Wilson's theorem (in case of confusion), concise: **Theorem (Wilson, 1770; implicit in Lagrange/Waring).** $n > 1$ is prime ⟺ $(n-1)! \equiv -1 \pmod n$. *Proof.* (⇐) If $(n−1)! ≡ −1 \pmod n$ and $d$ is a proper divisor of $n$ with $1 < d \le n−1$, then $d$ appears as a factor in $(n−1)!$, so $d \mid (n−1)!$; but $d \mid n \mid (n−1)! + 1$, hence $d \mid (n−1)!+1 − (n−1)! = 1$, contradiction. (⇒) Let p be prime. In the field $\mathbb{F}_p$, each $a \in \{1, \dots, p−1\}$ has a unique inverse $a^{−1}$ in the same set. Pairing elements with their inverses, elements with $a \ne a^{−1}$ contribute $a a^{−1} = 1$; the unpaired ones solve $a^2 \equiv 1 \pmod p$, i.e., $(a−1)(a+1) \equiv 0$, so $a = 1$ or $a = p−1$. Hence $(p−1)! \equiv 1 \cdot (p−1) \equiv −1 \pmod p$. ∎ Good. Also maybe define: $W_p$ values: $W_2 = 1, W_3 = 1, W_5 = 5, W_7 = 103, W_{11} = 329891$; Wilson primes are those with $p | W_p$. Formatting plan: - Brief intro paragraph: direct honest statement. - Section "What the conjecture says" — definitions + examples. - Section "I can't prove it — nobody can" — brief; status. Actually fold into intro + a "status" section. - Section "The strongest known 'argument': the heuristic" — with the computation. - Section "What is rigorously known" — bullets: Glaisher, searches, no-distribution-results, Carlitz contrast, cousins. - Section "Why a proof is out of reach with current methods" — short bullets. - Section "If you meant Wilson's theorem" — proof. - Closing offer (short list). Keep total moderate. Use LaTeX with $...$ (the renderer presumably supports it). Headers via ###. Bold sparingly. Double-check numbers: - 12! = 479001600 ✓; 12!+1 = 479001601 = 13² × 2834329 ✓ (verified above). - 4!+1 = 25 = 5² ✓. - W₇ = 103, and 7 ∤ 103 ✓ (7×14 = 98, +5). - Mertens constant 0.26149...✓. - log log 10^13: ln(10^13) = 13 × 2.302585 = 29.9336; ln(29.9336) = 3.399 ✓. +0.261 ≈ 3.66. - 4th prime expectation: solve log log x + 0.261 = 4 ⇒ log log x = 3.739 ⇒ log x = e^{3.739} = 42.05 ⇒ x = e^{42.05} ≈ 1.8 × 10^{18}. So "≈ 2 × 10^{18}". And crude: e^{e^4} = e^{54.6} ≈ 5.2 × 10^{23}. I'll say "somewhere in the range 10^{18}–10^{24}". - e^{e^{10}}: e^{10} ≈ 22026.5; e^{22026.5} = 10^{22026.5/2.3026} = 10^{9565}. ✓ ~10^{9565}. - Carlitz 1954 infinitely many irregular primes ✓. - Glaisher 1900 congruence ✓. - Goldberg 1953 found 563 ✓ (Kenneth Goldberg, SWAC). - Wieferich primes: 1093 (Meissner 1913), 3511 (Beeger 1922) ✓. - Wolstenholme primes 16843, 2124679 ✓ (Selfridge & Pollack 1964 found 16843; 2124679 later). - Wall–Sun–Sun primes: none known ✓ (searches to > 2^{64}? > 10^{17}? I'll just say none known). Statement of conjecture wording: "There are infinitely many Wilson primes." Sometimes also "the set of Wilson primes is infinite" — same. Also the related statement: for each k, the k-th Wilson prime exists. Also known as the "Wilson prime conjecture". Fine. One more possible inclusion: the expected number in [x, 2x] is ~1/log x → 0, so Wilson primes thin out; the count ~ log log x. Also mention: conjectured asymptotic #~ log log x isn't even a rigorous "conjecture" in the asymptotic sense (variance high: Poisson with mean log log x), but expectation → ∞. Also could mention equivalence: Wilson prime ⟺ $p^2 \mid (p-1)! + 1$ ⟺ $W_p \equiv 0 \pmod p$ ⟺ (Glaisher) $B_{p-1} + 1/p \equiv 1 \pmod p$. And note: 563: check that $B_{562} + 1/563 ≡ 1$? Not needed. Now write the response. Tone: respectful, clear, honest, helpful, not preachy. Start with the key fact: it's open. Avoid over-hedging; hedge where genuinely uncertain (search bound). Length target: maybe 600–900 words equivalent with formulas. It's a meaty topic; that's fine. Not too long — avoid bloating with every tangent (skip Lerch primes maybe one line; skip p-adic gamma or one line). Draft: --- **Short answer: this cannot be done — the Wilson prime conjecture is an open problem.** No proof (or disproof) is known; finding one would be major news in number theory. Rather than pretend otherwise, let me give you (1) a precise statement, (2) the strongest known argument in its favor, (3) what is rigorously known, and (4) why it resists proof — plus, in case it's what you wanted, a proof of Wilson's *theorem*, which is elementary. ### 1. The statement Wilson's theorem: $p$ prime ⟺ $(p-1)!\equiv -1 \pmod p$. Since $(p−1)!+1$ is then divisible by p, define Wilson quotient $W_p = \frac{(p-1)!+1}{p}$. **Wilson prime**: $p \mid W_p$, i.e. $(p−1)! \equiv −1 \pmod{p^2}$. Examples: $4!+1 = 25 = 5^2$; $12!+1 = 479001601 = 13^2\cdot 2834329$; $562!+1 \equiv 0 \pmod{563^2}$. Only three known: **5, 13, 563** (5,13 classical; 563 found by Goldberg in 1953 on SWAC). Searches (PrimeGrid, continuing) have checked everything up to order $10^{13}$ with no fourth — check current records for the exact bound. **Conjecture.** Infinitely many Wilson primes. ### 2. Why everyone believes it (the heuristic) No proof exists, but there's a persuasive model. If $W_p \bmod p$ behaves like a uniformly random residue (supported by computation and by analogy with Fermat quotients), then P(p Wilson) = 1/p, so expected # up to x = $\sum_{p\le x} 1/p = \log\log x + 0.2615... + o(1)$ (Mertens). This diverges ⇒ infinitely many predicted. The divergence is *very* slow: expected count by $10^{13}$ ≈ 3.7 vs 3 observed ✓; the 4th expected around $10^{18}$–$10^{24}$; the 10th around $10^{9565}$. So the data is fully consistent — but this is an argument from a randomness model, not a proof. (The same heuristic "proves" infinitude of Wieferich primes — 1093, 3511 — and Wolstenholme primes — 16843, 2124679 — none of which is proved either.) ### 3. What is rigorously known - Glaisher (1900): $W_p \equiv B_{p-1} + \frac1p - 1 \pmod p$ ($B$ = Bernoulli numbers). So Wilson primes ⟺ $p \mid \mathrm{num}(B_{p-1} + \frac1p - 1)$ — ties the problem to Bernoulli/irregularity theory. Carlitz (1954) did prove infinitely many irregular primes, but his argument needs the index to run over a growing range; the Wilson condition is a single diagonal index $a = p-1$, and the method doesn't reach it. - Empirically $W_p$ looks random mod p — but rigorously, not a single residue class of $W_p \bmod p$ is known to occur infinitely often (class 0 is our conjecture; class 1 = "Lerch primes": 3, 103, 839, 2237, open). - No unconditional upper bounds beyond the trivial: to current knowledge we cannot even prove that Wilson primes have density 0 among primes. Hmm, that last bullet — "cannot even prove density 0" — I'll keep with "to my knowledge". ### 4. Why it's hard Any proof of "infinitely many primes with property X" by current technology needs X to be governed by visible structure: congruence conditions (Dirichlet), values of L-functions/character sums, traces of Frobenius, sieve-dense sets, or monotone growth of some quantity. "$p^2$ divides $(p-1)!+1$" is a diagonal condition on the most composite of all objects; the correction term $W_p \bmod p$ is expected to be equidistributed but is controlled by no known analytic object. Cousins of the problem — infinitude of Wieferich primes, Wolstenholme primes, Wall–Sun–Sun primes, Mersenne primes — have resisted comparable efforts for a century+. A successful proof would need a genuinely new idea, which is exactly why the problem is famous. ### 5. In case you meant Wilson's theorem (provable!) Theorem + proof as drafted above. ### What I can help with next - prove Wilson's theorem in more detail / generalizations (Gauss's generalization to $\mathbb{Z}/n$); - prove Glaisher's congruence; - make the heuristic rigorous *inside the random model* (show expected count → ∞); - or, if you have a claimed proof in hand, I can help find where it breaks (a useful test: any correct argument must also explain why it doesn't prove infinitude of Mersenne or Wieferich primes, which are known to be open). --- That's a good shape. Trim wording. Also, in the bullet "No unconditional upper bounds...": phrase as "nothing nontrivial is known about the distribution: not even density zero is known." Also in section 3, maybe include the Lerch identity carefully? I decided to skip due to uncertainty. Could include the verified identity $\sum_{a=1}^{p-1} a^{p-1} \equiv p-1+pW_p \pmod{p^2}$ attributed hedged as "a classical congruence (going back to Lerch's work)"? I verified numerically for p=5,7. I believe this is right and is related to Lerch. Actually the known statement (Lerch 1905): $\sum_{a=1}^{p-1} a^{p-1} \equiv p - 1 + p\,W_p \pmod{p^2}$. Hmm wait, let me double check with p=3: $\sum a^2 = 1+4 = 5$. Formula: p−1 + pW₃ = 2 + 3×1 = 5 ✓ mod 9: 5 ≡ 5 ✓. p=11: sum a^10 mod 121 should = 10 + 11×329891 mod 121. 329891 mod 11: 3+2+9+8+9+1 = 32 → 3+2 = 5; digit-sum mod 11: 329891 = 11×29990 + 1 → 329891 ≡ 1? 11×29990 = 329890, remainder 1. So W_11 ≡ 1 mod 11. Then formula: 10 + 11×(329891) ≡ 10 + 11×1 = 21 mod 121. Direct: compute $\sum_{a=1}^{10} a^{10} \bmod 121$. By pairing a and 11−a: a^10 + (11−a)^10 ≡ a^10 + (−a + 11)^10 ≡ a^10 + a^10 − 10·11 a^9 ≡ 2a^10 − 110 a^9 (mod 121). Sum over a=1..10: but pairing (a, 11−a) pairs (1,10),(2,9),(3,8),(4,7),(5,6): sum over pairs of [2a^{10} − 110 a^9] for a=1..5. Compute a^10 mod 121 and a^9: a=1: 2(1) − 110(1) = 2 − 110 = −108 ≡ 13 (121−108). a=2: 2^9 = 512 ≡ 512 − 484 = 28; 2^10 = 56. 2(56) − 110(28) = 112 − 3080 = −2968. −2968 mod 121: 121×24 = 2904; −2968 + 3025 (121×25) = 57. So ≡ 57. a=3: 3^9 = 19683. mod 121: 121×162 = 19602; 19683−19602 = 81. 3^10 = 3×81 = 243 ≡ 243−242 = 1. 2(1) − 110(81) = 2 − 8910 = −8908. 121×73 = 8833; 121×74 = 8954. −8908 + 8954 = 46. ≡ 46. a=4: 4^9 = 262144. mod 121: 121×2166 = 262,086; 262144−262086 = 58. 4^10 = 4×58 = 232 ≡ 111. 2(111) − 110(58) = 222 − 6380 = −6158. 121×50 = 6050; 121×51 = 6171. −6158 + 6171 = 13. ≡ 13. a=5: 5^9 = 1953125. mod 121: 121×16141 = 1953061? 121×16000 = 1,936,000; 121×141 = 17,061; total 1,953,061. 1,953,125 − 1,953,061 = 64. 5^10 = 5×64 = 320 ≡ 320 − 242 = 78. 2(78) − 110(64) = 156 − 7040 = −6884. 121×57 = 6897. −6884 + 6897 = 13. ≡ 13. Sum: 13 + 57 + 46 + 13 + 13 = 142 ≡ 21 mod 121. ✓✓ Matches formula (21). So the identity $\sum_{a=1}^{p-1} a^{p-1} \equiv p - 1 + p W_p \pmod{p^2}$ is verified for p=3,5,7,11. I'm confident. Attribution to Lerch — I believe Lerch 1905 proved exactly this; and primes with $W_p ≡ 1 \pmod p$ are "Lerch primes". I'll include it as a fun fact with attribution hedged slightly ("a classical congruence of Lerch (1905)"). Actually to be safe I could write "going back to work of Lerch (1905)" — fine. Eh, is it needed? It's a nice concrete handle and shows the kind of "equivalent conditions" known. Keep it as one bullet. Wait, but should double check the claim "primes with $W_p ≡ 1$ are Lerch primes: 3, 103, 839, 2237". Check p=3: W₃=1 ≡ 1 ✓. p=103: is 103 a Lerch prime? I recall the Lerch primes are 3, 103, 839, 2237 (OEIS A111534?). I'm fairly confident. I'll mention without list maybe: "(e.g. 3, 103, 839, 2237)". Risk acceptable? If wrong it's a minor factual slip. I recall these four strongly — yes, Lerch primes: 3, 103, 839, 2237. Keep. Also double-check "Glaisher 1900" attribution for the Bernoulli congruence: Glaisher, "On the residues of ... powers" Q.J. Math 1900/1901 — yes, commonly cited as Glaisher (1900). ✓ One more consideration: the user might be a student who was asked "prove the Wilson prime conjecture" as a trick/awareness question, or a troll/test. Either way my response works. Also mention that 563 search: Goldberg found 563 as third Wilson prime in 1953 using SWAC ✓. Now: about the claim "not even density zero known". Actually hold on, let me reconsider — is there really no known result? There's a 2010s paper: "On the number of Wilson primes" by ... hmm, "P. Pollack"? Pollack has results on "the greatest prime factor of (p−1)!+1"? There's work on the largest prime factor of Wilson-related quantities: e.g., "the greatest prime factor of the product of ..." Also "Luca, ...: Wilson quotients are large"? There's a known result: $\max_{p \le x} W_p \bmod p$... no. There's a paper by "L. Toth"? Not recall. I'll keep hedged: "as far as I know, no nontrivial unconditional bound on their density is known." Also add a line: even proving that there are infinitely many primes p with $W_p \not\equiv 0$ — wait, that's trivial? Hmm: non-Wilson primes infinitude — as discussed, unclear! Let me test whether trivial: We know only 3 Wilson primes ≤ 10^13 but that's computational, not proof. Is there a slick proof that infinitely many primes fail $(p-1)! ≡ −1 \pmod {p^2}$? Hmm... Suppose not: all sufficiently large primes are Wilson primes. Consider... p and 2p−... I can't see any. I'll include this as a striking bullet: "we cannot even rule out (unconditionally) that all but finitely many primes are Wilson primes" — hmm, this is a strong claim about the literature. Risky but I believe it's true and it's memorable. Actually, hmm, wait. Let me think once more about whether the real Lerch congruence could rule it out. If the true Lerch congruence is (per my failed tests, it's something else) — since I couldn't validate the recalled form, maybe the actual form is: $$\#\{p \le x : p \text{ Wilson prime}\} \equiv \sum_{p \le x} \frac{(p-1)!+1}{p} \pmod{p_{\text{next}}}$$ — tested false. OR maybe the identity is about $\sum_{p\le x} W_p \equiv \#\{...\} \pmod {p}$ where the modulus is EACH prime in the sum?? Like: for each prime $q \le x$: $\sum_{p \le x} W_p \equiv$ (number of Wilson primes ≤ x... ) hmm mod q? Test: q=5, sum over p ≤ 7: 110 ≡ 0 mod 5. # Wilson ≤ 7 = 1. 0≠1. Fail. Maybe it involves $\binom{p-1}{k}$... The real Lerch theorem (I now actually recall reading it in Crandall–Pomerance §3? or in "Concrete Mathematics"?) — Lerch: $$\sum_{\substack{p \le x \\ p \text{ prime}}} \frac{(p-1)!+1}{p} \equiv \#\{p \le x: p \text{ Wilson}\} \pmod{p^*}$$ where $p^*$ is the least prime > x? tested false. Since I can't verify, OMIT Lerch congruence entirely; keep the verified $\sum a^{p-1}$ identity with hedge "due to Lerch". Hmm, actually let me reconsider whether even that attribution is safe. The identity: for prime p, $\sum_{a=1}^{p-1} a^{p-1} \equiv p - 1 + p W_p \pmod{p^2}$. Verified p = 3,5,7,11 numerically. This is very likely classical (it appears in papers on Lerch primes; the Lerch quotient $l_p = \frac{\sum_{a=1}^{p-1} a^{p-1} - p}{p}$ and $l_p \equiv W_p - 1 \pmod p$). Yes: $l_p = \frac{\sum a^{p-1} - p}{p} \equiv W_p − 1 \pmod p$. And Lerch primes: $l_p ≡ 0$ i.e. $W_p ≡ 1$. Consistent. Attribution to Lerch (1905) — I'm ~90% sure. I'll write "a classical congruence (Lerch, 1905)". Hmm OK wait, I realize I should double-check the claim "Goldberg, 1953, SWAC". Kenneth Goldberg, "A search for Wilson primes" (1953?) using SWAC under D.H. Lehmer — I'm fairly confident: yes, K. Goldberg found 563 with SWAC in 1953. Keep but phrase lightly "(found by K. Goldberg in 1953 on the early SWAC computer)". Search bound: I'll say: "computational searches (currently maintained by PrimeGrid) have verified there is no fourth Wilson prime up to roughly $10^{13}$" + "I'd double-check the current bound" — hmm, phrasing "roughly" plus "as of my knowledge" is fine. Let me phrase: "as of the searches I'm aware of, all primes up to at least about $10^{13}$ have been checked". Also: the first three: "5, 13, 563 (OEIS A007540)". Could include OEIS ref — fine, adds credibility. A007540: Wilson numbers? Hmm, A007540 is "Wilson primes: primes p such that (p−1)!+1 ≡ 0 mod p²: 5, 13, 563". I believe A007540 is indeed Wilson primes. Keep or drop? Drop OEIS ID to be safe, or say "(OEIS: search 'Wilson primes')". I'll just say only three are known. Regarding "562!+1 ≡ 0 (mod 563²)": 563² = 316,969. The statement is standard. ✓ Now, also perhaps note the sequence: expected count up to x ~ log log x: up to 10^13 ≈ 3.7. Also up to 10^100: log log 10^100 = ln(230) ≈ 5.44 + 0.26 ≈ 5.7. So even up to 10^{100} expect ~6 Wilson primes. Fun. The final offer section — keep tight. Also I should be careful to not make the response feel like a lecture to a troll; keep it warm and precise. Also the instruction: no need to mention my guidelines etc. Word budget: aim ≤ ~850 words. Use headers (##/###) and bullets, LaTeX. Let me write it. Also start: "I can't — and no one can (yet)." Good hook. Then "Here's an honest tour of the problem..." Write final. Check formatting guidance: This is a technical math chat; markdown with headers, LaTeX, bullets appropriate. Length medium-long is fine given the ask. Avoid excessive nesting. One more sanity check on Mertens constant name: "Meissel–Mertens constant" $M \approx 0.2615$. ✓ Also double-check 5, 13 classical: 13 known by... whoever; safe to say "5 and 13 were long known". Glaisher knew? Probably; just say "5 and 13 have been known since the 19th century" — hmm, is that certain? The notion of Wilson primes existed by 1900 (Glaisher computed W_p mod p for p < ... and observed 5, 13). I'd say "5 and 13 were known to 19th-century computers of the factorial residues; 563 found by K. Goldberg in 1953". Fine. Also, note: could there be any Wilson prime between 563 and search bound? No — that's the point of searches. Final structure: 1. **Opening** (3–4 sentences): can't prove; open; would be major result; here's the honest state of the art + best "argument". 2. **The statement** — Wilson quotient definition, Wilson prime, examples with the two explicit factorizations, the conjecture, search status. 3. **The heuristic** (the best known 'argument') — model, Mertens, divergence, numeric agreement, 4th prime prediction, contrast note (model ≠ proof; same model "proves" open problems). 4. **Rigorously known** — bullets: Glaisher congruence & Bernoulli reformulation; Lerch power-sum identity; Carlitz irregular primes contrast; no residue class known infinite; no density control; (computational bounds). 5. **Why hard** — short paragraph: diagonal condition, factorial, no L-function/sieve handle; cousins open; new idea needed. 6. **If you meant Wilson's theorem** — statement + proof (concise, complete). 7. **Offers** — 3–4 bullets. Go. ## Short answer I can't — and neither can anyone. The **Wilson prime conjecture is a genuinely open problem** in number theory. No proof (or disproof) that infinitely many Wilson primes exist is known; a proof would be a major result. Rather than fabricate one, let me give you an honest tour: the precise statement, the strongest known *argument* for it, what is rigorously known, why it resists proof — and, in case this is what you actually wanted, a proof of **Wilson's theorem**, which *is* elementary. ## 1. The statement By Wilson's theorem, $p$ is prime iff $(p-1)! \equiv -1 \pmod p$. The **Wilson quotient** is $$W_p = \frac{(p-1)! + 1}{p} \in \mathbb{Z}.$$ A **Wilson prime** is a prime with $p \mid W_p$, i.e. $(p-1)! \equiv -1 \pmod{p^2}$. For example: $$4!+1 = 25 = 5^2, \qquad 12!+1 = 479{,}001{,}601 = 13^2 \cdot 2{,}834{,}329.$$ Exactly three Wilson primes are known: **5, 13, 563** (563 was found by K. Goldberg in 1953 on the SWAC computer). Large searches (maintained, as I understand it, by PrimeGrid and predecessors) have covered all primes up to roughly $10^{13}$ without a fourth example — do check current records for the exact bound. **Conjecture.** *There are infinitely many Wilson primes.* ## 2. Why everyone believes it (the best known "argument") If $W_p \bmod p$ behaves like a uniformly random residue mod $p$ (supported by computation and analogy with Fermat quotients), then the probability that $p$ is a Wilson prime is $1/p$, so the expected count up to $x$ is $$\sum_{p \le x} \frac{1}{p} = \log\log x + 0.2615\ldots + o(1),$$ which diverges — hence infinitely many are predicted. The divergence is *extremely* slow: the expected count by $10^{13}$ is $\approx 3.7$ versus $3$ observed ✓; the fourth example is expected around $10^{18}$–$10^{24}$; the tenth around $10^{9565}$. The data fits beautifully — but this is an argument from a randomness model, **not a proof**. (The same heuristic "proves" infinitude of Wieferich primes — only $1093, 3511$ known — and Wolstenholme primes — $16843, 2124679$ — none of which is proved either.) ## 3. What is rigorously known - **Glaisher (1900):** $W_p \equiv B_{p-1} + \tfrac1p - 1 \pmod p$, where $B_n$ are Bernoulli numbers. So Wilson primes are exactly primes with $p \mid \mathrm{num}\!\left(B_{p-1} + \tfrac1p - 1\right)$, tying the problem to irregularity theory. Carlitz (1954) *did* prove infinitely many irregular primes, but his growth argument needs the Bernoulli index to range widely; the Wilson condition is a single "diagonal" index $a = p-1$, out of reach of that method. - **A classical congruence (Lerch, 1905):** $\displaystyle\sum_{a=1}^{p-1} a^{p-1} \equiv p - 1 + pW_p \pmod{p^2}$ — a different handle on $W_p$, still yielding nothing about infinitude. - **Not a single residue class of $W_p \bmod p$ is known to be attained infinitely often.** Class $0$ is the conjecture itself; class $1$ defines the "Lerch primes" ($3, 103, 839, 2237$), whose infinitude is also open. - **No nontrivial unconditional distribution results.** To my knowledge we cannot even prove the Wilson primes have density zero among the primes. ## 4. Why it's out of reach Proving "infinitely many primes with property $X$" by current technology requires $X$ to be governed by visible structure: congruence conditions (Dirichlet), $L$-values, character sums, traces of Frobenius, sieve-friendly densities, or monotone growth. The condition $p^2 \mid (p-1)!+1$ is a diagonal condition on the most composite of all objects, and $W_p \bmod p$ is controlled by no known analytic object. Its cousins — infinitude of Wieferich, Wolstenholme, Wall–Sun–Sun, or Mersenne primes — have resisted comparable efforts for over a century. A proof needs a genuinely new idea. ## 5. In case you meant *Wilson's theorem* (this one is provable) **Theorem.** For $n > 1$: $n$ is prime $\iff (n-1)! \equiv -1 \pmod n$. **Proof.** ($\Leftarrow$) Suppose $(n-1)! \equiv -1 \pmod n$ and $d$ is a divisor with $1 < d \le n-1$. Then $d$ appears among the factors of $(n-1)!$, so $d \mid (n-1)!$; also $d \mid (n-1)! + 1$. Hence $d \mid 1$, a contradiction. ($\Rightarrow$) Let $p$ be prime. In $\mathbb{F}_p$, every $a \in \{1, \dots, p-1\}$ has a unique inverse $a^{-1}$ in that set. Pairing $a$ with $a^{-1}$, each pair contributes $aa^{-1} \equiv 1$; the unpaired elements satisfy $a^2 \equiv 1$, i.e. $(a-1)(a+1) \equiv 0$, so $a \equiv \pm1$. Thus $(p-1)! \equiv 1 \cdot (-1) \equiv -1 \pmod p$. $\blacksquare$ ## What I can help with next - A detailed proof of Wilson's theorem or Gauss's generalization; - A proof of Glaisher's congruence $W_p \equiv B_{p-1} + \tfrac1p - 1 \pmod p$; - Making the heuristic rigorous *within the random model* (showing the expected count really tends to infinity); - If you have a claimed proof in hand, I can help locate where it breaks — a useful sanity check is that any correct argument must also explain why it doesn't instantly prove the infinitude of Mersenne or Wieferich primes, which are known to be open.

Please prove the Wilson prime conjecture