Obviously you mean n>m for arbitrary m right? Bomb

orl wrote: Prove that there exists an integer $ n$, $ n\geq 2002$, and $ n$ distinct positive integers $ a_1,a_2,\ldots,a_n$ such that the number $ N = a_1^2a_2^2\cdots a_n^2 - 4(a_1^2 + a_2^2 + \cdots + a_n^2)$ is a perfect square. Does anyone have solution?

Let $m \geq 3$ be a positive integer. We first claim that if there exists some positive integer $S$ and distinct integers $a_1, a_2, \ldots, a_m$ such that \[ a_1 a_2 \cdots a_m = a_1^2 + a_2^2 + \cdots + a_m^2 + S, \] then there exists some positive integer $a_{m+1}$ distinct from each of $a_1, a_2, \ldots, a_m$ such that \[ a_1 a_2 \cdots a_m a_{m+1} = a_1^2 + a_2^2 + \cdots + a_m^2 + a_{m+1}^2 + S-1. \] Indeed, if we take $a_{m+1} = a_1 a_2 \cdots a_m - 1 > \max(a_1, a_2, \ldots, a_m)$, we get \begin{align*} a_1 a_2 \cdots a_m a_{m+1} &= (a_1 a_2 \cdots a_m)^2 - a_1 a_2 \cdots a_m = (a_1 a_2 \cdots a_m - 1)^2 + a_1 a_2 \cdots a_m - 1 \\ &= (a_1 a_2 \cdots a_m - 1)^2 + (a_1^2 + a_2^2 + \cdots + a_m^2 + S) - 1 = a_1^2 + a_2^2 + \cdots + a_m^2 + a_{m+1}^2 + S-1. \end{align*} We now observe that we can apply the claim repeatedly to find some $a_{m+1}, a_{m+2}, \ldots, a_{m+S-1}$ such that \[ a_1 a_2 \cdots a_{m+S-1} = a_1^2 + a_2^2 + \cdots + a_{m+S-1}^2 + 1. \] Now define $a_i = i$ for $1 \leq i \leq 2002$ and define \[ S = a_1 a_2 \cdots a_{2002} - (a_1^2 + a_2^2 + \cdots + a_{2002}^2) = 2002! - (1^2 + 2^2 + \cdots + 2002^2) > 0. \] Let $n=2002+S-1 \geq 2002$. By the above claim, we can find $a_{2003}, a_{2004}, \ldots, a_n$ with \[ a_1 a_2 \cdots a_n = a_1^2 + a_2^2 + \cdots + a_{m+S-1}^2 + 1. \] It follows that \[ a_1^2 a_2^2 \cdots a_n^2 - 4(a_1^2 + a_2^2 + \cdots + a_n^2) = a_1^2 a_2^2 \cdots a_n^2 - 4a_1 a_2 \cdots a_n + 4 = (a_1 a_2 \cdots a_n - 2)^2 \] is a perfect square, so our construction is complete.

We will find for $n = 3a^2 - 14 \ (a\ge 5)$, $a_1,...,a_n$ such that $N = (\prod_{i=1}^n a_i - 2a)^2$. Equating this with the original expression for $N$ we end up with the equation $$\sum_{i=1}^n a_i^2 - 2a\prod_{i=1}^n x_i + 4a^2 = 0. \qquad \qquad (*)$$Construct an infinite sequence of positive integers as follows: let $x_1 = ... = x_{n-2} = 1, x_{n-1} = 4, x_n = a$, and $$ \forall m \in \mathbb N, x_{n+m} = 2a\prod_{j=1}^{n-1} x_{j+m} - x_m.$$Induction shows that $(x_{n+m})_{m\in \mathbb N}$ is strictly increasing and every $n$ consecutive terms of $(x_i)_{i\in \mathbb N}$ is a solution to $(*)$ (ignoring "pairwise distinct" constraint). However as $x_{n+1},...,x_{2n}$ are pairwise distinct, we can take $a_i = x_{n+i}$ to satisfy the original problem.