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 has this property that is a perfect square for all Prove that for some

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- Oct 9th 2009, 05:56 AMNonCommAlgTough & 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 has this property that is a perfect square for all Prove that for some - Oct 16th 2009, 04:48 PMeeyore
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!

- Oct 19th 2009, 01:08 PMOpalg
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 4*a*, 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?? (Happy) - Oct 20th 2009, 08:01 PMSampras
has to be reducible. Maybe use proof by contradiction. Suppose was not a "perfect square."

- Oct 21st 2009, 12:39 AMNonCommAlg
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.