GPT 6 Astra Ultra attempt on Math-5789
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Erdős Problem 646: even prime exponents in factorials
Statement.
For any finite list of distinct primes $p_1,\ldots,p_k$, are there infinitely many positive integers $n$ for which all $v_{p_i}(n!)$ are even?Known theorem and reduction.
The answer is yes. The external input is Daniel Berend's theorem: for every $r\geq1$, infinitely many $n$ have even factorial valuations at each of the first $r$ primes. See On the parity of exponents in the factorization of $n!$, Journal of Number Theory 64 $1997$, 13--19, https://doi.org/10.1006/jnth.1997.2106. The author's university publication record states this theorem explicitly: https://cris.bgu.ac.il/en/publications/on-the-parity-of-exponents-in-the-factorization-of-n-4/. This is the established resolution of https://www.erdosproblems.com/646.To check that the theorem covers arbitrary finite prime lists, take $r=\pi(\max_i p_i)$. Every prescribed $p_i$ lies among the first $r$ primes. Each of the infinitely many integers supplied by Berend therefore meets every requested parity condition. No independence assumption between different primes is being made. Berend's simultaneous recurrence theorem is used externally, rather than claimed as proved by this reduction.
Elementary verification and a complete one-prime construction.
Counting the multiples of $p,p^2,\ldots$ in a factorial gives Legendre's identity \[ v_p(n!)=\sum_{j\geq1}\left\lfloor\frac n{p^j}\right\rfloor =\frac{n-s_p(n)}{p-1}, \] where $s_p(n)$ is the sum of the base-$p$ digits. The second equality follows by expanding $n=\sum_i d_ip^i$ and summing the resulting geometric series. For $n=p^a-1$, all $a$ digits equal $p-1$, and hence \[ v_p((p^a-1)!)=\frac{p^a-1}{p-1}-a =\sum_{j=0}^{a-1}p^j-a. \] If $p$ is odd, every summand is odd, so this valuation is even for every $a\geq1$. If $p=2$, it equals $2^a-1-a$, which is even whenever $a$ is odd. These give explicit infinite families for the one-prime case.There is also a useful reformulation of the simultaneous problem. Let $S=\{p_1,\ldots,p_k\}$, and remove from $n!$ all prime powers whose primes do not belong to $S$. The resulting $S$-part is $\prod_{p\in S}p^{v_p(n!)}$, and unique factorisation says that it is a square exactly when all the required valuations are even. The one-prime families above do not establish that simultaneous condition; this is precisely the role of Berend's theorem.