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Math Help - Tough & Beautiful!

  1. #1
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    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 p(x) \in \mathbb{Z}[x] has this property that p(n) is a perfect square for all n \in \mathbb{Z}. Prove that p(x)=(q(x))^2, for some q(x) \in \mathbb{Z}[x].
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  2. #2
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    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!
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  3. #3
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    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 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, 60x^2-60x+1 is a square when x = –2, –1, 0, 1, 2 and 3.)

    Let 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

    \Bigl(\sqrt a x+y + \frac b{2\sqrt a}\Bigr)\Bigl(\sqrt a x-y + \frac b{2\sqrt a}\Bigr) = \frac{b^2-4ac}{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 \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 \sqrt a x will be densely distributed over the unit interval as x\to\infty, so they cannot all be close to the fractional part of -\tfrac b{2\sqrt a} and hence \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, a=k^2 say.

    Now if we multiply (*) by 4a, we can write it as

    (2k^2x+2ky+b)(2k^2x-2ky+b) = b^2-4ac.\qquad(**)

    If b^2-4ac\ne0 then it only has finitely many factors, so (**) has only finitely many integer solutions. Since that is not the case, we must have b^2=4ac.

    Going back to the polynomial p(x) = ax^2+bx+c, we know that p(0)=c is a square, say c=l^2. Then b^2=4ac=4k^2l^2, so b=\pm2kl. Finally, 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??
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  4. #4
    Senior Member Sampras's Avatar
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     p(x) has to be reducible. Maybe use proof by contradiction. Suppose  p(x) was not a "perfect square."
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  5. #5
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    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 P(x) be a counter-example. then:

    1) we may assume that \deg P(x) \geq 2 and that P(x) has no repeated roots. so if \Delta is the discriminant of P(x), then \Delta \neq 0.

    2) for any (positive) integer m there exist a prime number p and an integer n such that p \mid P(n) and p \nmid m. (why?)

    3) for m=\Delta let p and n be as mentioned in 2). since P(n) is a perfect square, we have p^2 \mid P(n). thus p^2 \mid P(n+p) - pP'(n).

    4) p \mid P(n) and p \nmid \Delta implies that p \nmid P'(n). (why?) therefore p^2 \nmid P(n+p), by 3), and so P(n+p) is not a perfect square. Q.E.D.
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