Determine all integers $1 \le m, 1 \le n \le 2009$, for which \begin{align*} \prod_{i=1}^n \left( i^3 +1 \right) = m^2 \end{align*}
Problem
Source: Balkan MO ShortList 2009 N3
Tags: number theory
kaede
07.04.2020 19:51
Assume that $P=\prod ^{n}_{i=1}\left( i^{3} +1\right)$ is a perfect square for some $n\in \mathbb{N}$ with $n\leq 2009$.
Note that $\prod ^{n}_{i=1}\left( i^{3} +1\right) =( n+1) !\cdotp \prod ^{n}_{i=1}\left( i^{2} -i+1\right)$.
So if $q$ is a prime divisor of $P$ and $q\equiv 2\bmod 3$, then $2\mid \nu _{q}(( n+1) !)$.
Since $2\mid P$, we have $2\mid \nu _{2}(( n+1) !)$, which implies $n\geq 5$.
Since $n\geq 4$, we have $5\mid P$, which implies $\nu _{5}(( n+1) !) \geq 2$ and so $n\geq 9$.
If $n=9$, then $\nu _{3}( P) =7$, which is impossible.
So we may assume, henceforth, $n\geq 10$.
Since $n\geq 10$, we have $11\mid P$, which implies $\nu _{11}(( n+1) !) \geq 2$ and so $n\geq 21$.
Since $n\geq 16$, we have $17\mid P$, which implies $\nu _{17}(( n+1) !) \geq 2$ and so $n\geq 33$.
Since $n\geq 28$, we have $29\mid P$, which implies $\nu _{29}(( n+1) !) \geq 2$ and so $n\geq 57$.
Note that $51^{2} -51+1=2551$ is a prime number.
Let $p=2551$.
If $p\mid ( n+1) !$, then $n\geq 2550$, which contradicts that $n\leq 2009$.
So we must have $p\mid j^{2} -j+1$ for some $j\in \mathbb{N}$ with $52\leq j\leq 2009$.
Note that $51^{2} -51+1\equiv j^{2} -j+1\equiv ( p-50)^{2} -( p-50) +1\equiv 0\bmod p$.
It follows that the number of solutions of the congruence $x^{2} -x+1\equiv 0\bmod p$ is at least $3$.
This is clearly impossible.
In fact, $\prod ^{n}_{i=1}\left( i^{3} +1\right)$ is non-perfect square unconditionally.
This follows almost directly from the lemma below :
Lemma
Let $n$ be positive integer greater than $10$.
There exist a prime $p$ such that $p\equiv 2\bmod 3$ and $n/2< p\leq n$.
Proof
Let $( a,q)$ be a pair of positive integers with $\gcd( a,q) =1$.
For each $x\in \mathbb{R}$, let $\pi ( x;\ q,a) =\#\{r\in \mathbb{P} |\ r\leq x,\ r\equiv a\bmod q\}$
In fact, if $x\geq 450$, then $\frac{x}{2\log( x)} < \pi ( x;\ 3,2) < \frac{x}{2\log( x)}\left( 1+\frac{5}{2\log( x)}\right)$
See Corollary 1.6 at http://www.math.ubc.ca/~andrewr/pubs/EBPAP.pdf
In particular, $\pi ( x;\ 3,2) -\pi ( x/2;\ 3,2) \geq 1$ when $x\geq 450$.
For $11\leq x< 450$, we can manually check that $\pi ( x;\ 3,2) -\pi ( x/2;\ 3,2) \geq 1$.
$\blacksquare$
Chevrolet23
18.06.2022 17:33
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