
Tough & Beautiful!
I don't know that much about number theory but those who do might have seen this problem. See if you can solve it:
Suppose $\displaystyle p(x) \in \mathbb{Z}[x]$ has this property that $\displaystyle p(n)$ is a perfect square for all $\displaystyle n \in \mathbb{Z}.$ Prove that $\displaystyle p(x)=(q(x))^2,$ for some $\displaystyle q(x) \in \mathbb{Z}[x].$

I've been thinking about this problem for days! I really want to solve it. I had a fledgling idea that used Eisenstein's criterion but I convinced myself that I was misusing the theorem. Still working on it though!

I have been thinking about this problem on and off for a while, without getting very far. Even the very simplest case, when the polynomial is quadratic, seems less than obvious.
So suppose that $\displaystyle p(x) = ax^2 + bx + c$ is a square for all integers x. (Notice in passing that p(x) can take quite a lot of square values without being the square of a linear polynomial. For example, $\displaystyle 60x^260x+1$ is a square when x = –2, –1, 0, 1, 2 and 3.)
Let $\displaystyle y^2 = ax^2+bx+c$. Clearly a>0, otherwise the quadratic would become negative for large x. We can factorise this equation over the reals as
$\displaystyle \Bigl(\sqrt a x+y + \frac b{2\sqrt a}\Bigr)\Bigl(\sqrt a xy + \frac b{2\sqrt a}\Bigr) = \frac{b^24ac}{4a}.\qquad(*)$
If x is very large then so is y, so the first of the factors on the left side of (*) will be very large, and the second factor will be very small. Thus $\displaystyle \sqrt a x + \tfrac b{2\sqrt a}$ will be close to the integer y. If a is not a perfect square then the values of the fractional part of $\displaystyle \sqrt a x$ will be densely distributed over the unit interval as $\displaystyle x\to\infty$, so they cannot all be close to the fractional part of $\displaystyle \tfrac b{2\sqrt a}$ and hence $\displaystyle \sqrt a x + \tfrac b{2\sqrt a}$ cannot always be close to an integer. That is a contradiction, and it shows that a must be a square, $\displaystyle a=k^2$ say.
Now if we multiply (*) by 4a, we can write it as
$\displaystyle (2k^2x+2ky+b)(2k^2x2ky+b) = b^24ac.\qquad(**)$
If $\displaystyle b^24ac\ne0$ then it only has finitely many factors, so (**) has only finitely many integer solutions. Since that is not the case, we must have $\displaystyle b^2=4ac$.
Going back to the polynomial $\displaystyle p(x) = ax^2+bx+c$, we know that $\displaystyle p(0)=c$ is a square, say $\displaystyle c=l^2$. Then $\displaystyle b^2=4ac=4k^2l^2$, so $\displaystyle b=\pm2kl$. Finally, $\displaystyle p(x) = k^2x^2\pm2klx+l^2 = (kx\pm l)^2$, so p(x) is the square of an integer polynomial, as required.
That doesn't look like a hopeful approach for generalising to the case of higher degree polynomials. Perhaps it's time for NonCommAlg to shed a bit of light on this?? (Happy)

$\displaystyle p(x) $ has to be reducible. Maybe use proof by contradiction. Suppose $\displaystyle p(x) $ was not a "perfect square."

as far as i know, there are 3 or 4 proofs of this problem. here's a sketch of the "easiest" one:
suppose the claim is not true and let $\displaystyle P(x)$ be a counterexample. then:
1) we may assume that $\displaystyle \deg P(x) \geq 2$ and that $\displaystyle P(x)$ has no repeated roots. so if $\displaystyle \Delta$ is the discriminant of $\displaystyle P(x),$ then $\displaystyle \Delta \neq 0.$
2) for any (positive) integer $\displaystyle m$ there exist a prime number $\displaystyle p$ and an integer $\displaystyle n$ such that $\displaystyle p \mid P(n)$ and $\displaystyle p \nmid m.$ (why?)
3) for $\displaystyle m=\Delta$ let $\displaystyle p$ and $\displaystyle n$ be as mentioned in 2). since $\displaystyle P(n)$ is a perfect square, we have $\displaystyle p^2 \mid P(n).$ thus $\displaystyle p^2 \mid P(n+p)  pP'(n).$
4) $\displaystyle p \mid P(n)$ and $\displaystyle p \nmid \Delta$ implies that $\displaystyle p \nmid P'(n).$ (why?) therefore $\displaystyle p^2 \nmid P(n+p),$ by 3), and so $\displaystyle P(n+p)$ is not a perfect square. Q.E.D.