Ngtjyty

In essence, I want to prove that the product of any two distinct elements in the set ${n^2, n^2+1, ... , (n+1)^2-1}$ is never a perfect square for a positive integers $n$. I have no idea on how to prove it, but I've also yet to find a counterexample to this statement. Can anyone help?

First, note that $n^2$ can't be one of the elements as the other element would also need to be a perfect square for the product to be a perfect square. As gnasher729 commented to the answer, this is why $left(n + 1right)^2$ is not included.

Assume there are $2$ such elements, $n^2 + a$ and $n^2 + b$, with $a neq b$, $1 le a, b le 2n$ and, WLOG, $a lt b$. Thus, consider their product to be a perfect square of some integer $p$, i.e.,

$$left(n^2 + aright)left(n^2 + bright) = p^2 tag{1}label{eq1}$$

As $n^2 + a lt p lt n^2 + b$, let

$$p = n^2 + c tag{2}label{eq2}$$

for some $a lt c lt b$. Thus, for some positive integers $d$ and $e$, we have that

$$a = c - d tag{3}label{eq3}$$
$$b = c + e tag{4}label{eq4}$$

Substitute eqref{eq2}, eqref{eq3} and eqref{eq4} into eqref{eq1} to get

$$left(n^2 + left(c - dright)right)left(n^2 + left(c + eright)right) = left(n^2 + cright)^2 tag{5}label{eq5}$$

Expanding both sides gives

$$n^4 + 2cn^2 + left(e - dright)n^2 + c^2 + cleft(e - dright) - ed = n^4 + 2cn^2 + c^2 tag{6}label{eq6}$$

Removing the common terms on both sides and moving the remaining terms, apart from the $n^2$ one, to the right gives

$$left(e - dright)n^2 = -cleft(e - dright) + ed tag{7}label{eq7}$$

Note that $e le d$ won't work because the LHS would be become non-positive but the RHS would become positive. Thus, consider $e gt d$, i.e., let

$$e = d + m, text{ with } m ge 1 tag{8}label{eq8}$$

Thus,

$$e + d lt 2n Rightarrow 2d + m lt 2n Rightarrow d lt n - frac{m}{2} tag{9}label{eq9}$$

Also,

$$ed = left(d + mright)d lt left(n + frac{m}{2}right)left(n - frac{m}{2}right) = n^2 - frac{m^2}{4} lt n^2 tag{10}label{eq10}$$

As such, the RHS of eqref{eq7} is $lt n^2$, so it can't be true.

For any two numbers $n^2+a, n^2+b; 0<a<b<(2n+1)$, their product will satisfy $n^4<n^4+(a+b)n^2+ab<(n^2+2n+1)^2$.

All of the squares between $n^4$ and $(n^2+2n+1)^2$ will have the form $(n^2+m)^2=n^4+2mn^2+m^2; 1le mle 2n$

If $n^4+(a+b)n^2+ab$ is a perfect square, it will be one of the squares between $n^4$ and $(n^2+2n+1)^2$ and hence equal to $n^4+2mn^2+m^2$ for some $m$.

Thus $(a+b)=2m; ab=m^2$ for some $m$

Rearranging, we get $m^2=frac{a^2+2ab+b^2}{4}=ab=frac{4ab}{4}$, or $a^2+b^2=2ab$.

This implies both $amid b$ and $bmid a$, meaning $b=a$ and the numbers being multiplied to obtain a perfect square are not distinct.