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 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, is a square when x = –2, –1, 0, 1, 2 and 3.)
Let . Clearly a>0, otherwise the quadratic would become negative for large x. We can factorise this equation over the reals as
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 will be close to the integer y. If a is not a perfect square then the values of the fractional part of will be densely distributed over the unit interval as , so they cannot all be close to the fractional part of and hence cannot always be close to an integer. That is a contradiction, and it shows that a must be a square, say.
Now if we multiply (*) by 4a, we can write it as
If then it only has finitely many factors, so (**) has only finitely many integer solutions. Since that is not the case, we must have .
Going back to the polynomial , we know that is a square, say . Then , so . Finally, , 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??
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 be a counter-example. then:
1) we may assume that and that has no repeated roots. so if is the discriminant of then
2) for any (positive) integer there exist a prime number and an integer such that and (why?)
3) for let and be as mentioned in 2). since is a perfect square, we have thus
4) and implies that (why?) therefore by 3), and so is not a perfect square. Q.E.D.