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$p(x)=ax^2+bx+c$ sends all integers to perfect squares. I wonder whether $p(x)$ is prefect square? I only know that $a$ & $b$ should be half integers. But how to use perfect square numbers here?

Kirby Lee
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    Why should $a$ and $b$ be half integers? Can they not be integers? – Arthur Sep 01 '16 at 14:53
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    Please don't leave the problem statement out of the body of the Question, relying only on the title to bear this burden. This invites confusion as the Reader tries to connect the incomplete statements in the Question body. – hardmath Sep 01 '16 at 14:58
  • The title isn't a grammatically correct / punctuated sentence. Sometimes the question can be inferred but in this case I'm confused. – Matthew Leingang Sep 01 '16 at 15:04
  • @Arthur Integers are half-integers. I think that Kirby is using the fact that a quadratic polynomial that takes integer values at integers must have half-integer coefficients and an integer constant term. – Andrew Dudzik Sep 01 '16 at 15:10
  • @Slade Google disagrees with you. – Arthur Sep 01 '16 at 15:13
  • @Arthur I guess that my usage is not as common as I thought. In any case, Kirby clearly means "an integer divided by $2$". – Andrew Dudzik Sep 01 '16 at 15:16
  • @Slade I guess my first comment could equally well have been "I see you're using that word. I don't think it means what you think it means." It could be both. My point is, when the OP writes "I know only that $a$ & $b$ should be half integers", he's either misunderstood the result about polynomials with integer values, or the meaning of "half integer". I don't know which, but as it stands it's incorrect, and my original content addressed that in one way (of admittedly many possible). – Arthur Sep 01 '16 at 15:23
  • More generally see here. $\ \ $ – Bill Dubuque Dec 16 '24 at 19:15

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First, let's establish that $a,b,c$ are integers. As observed, we have $2a,2b,c\in\mathbb{Z}$ from the fact that $p$ takes integer values.

Since $p(4)=16a+4b+c$ is a square, it is $0$ or $1$ modulo $4$. But $c$ is a square, and therefore also $0$ or $1$ modulo $4$, and $4b$ is an even integer, therefore $4b\equiv 0 \pmod{4}$ and $b\in\mathbb{Z}$.

Similarly, $p(2)=4a+2b+c$ is a square, and $2b+c$ is $0$ or $1$ modulo $4$, so $a\in\mathbb{Z}$.

Now we can follow the argument here: choose a prime $q$ such that $q \mid p(n)$ for some $n\in\mathbb{Z}$, large enough so that $q$ does not divide the resultant $-a(b^2-4ac)$ of $p(x)$ and $p'(x)$.

Then $q^2 \mid p(n)$, because $p(n)$ is a square. But $p(n+q)\equiv p(n)+q p'(n) \pmod{q^2}$. So $q\mid p(n+q)$, therefore $q^2\mid p(n+q)$, and $q\mid p'(n)$, so that $p$ and $p'$ have a common root modulo $q$, contradicting the assumption that $q$ does not divide the resultant.

The only way this could have happened is if the resultant were zero, which happens precisely when $p(x)$ has a double root, i.e. is the square of a linear polynomial.

Apparently it is an open problem in algebraic geometry to determine how many consecutive square values a quadratic may take. But here is a quadratic that takes $8$ consecutive square values, at $x=-1,0,1,2,3,4,5,6$:

$$p(x) = -420x^2 + 2100x + 2809$$

Andrew Dudzik
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  • Thank you so much. I'm reading the article linked above. It's of great help and seems hard to digest. Orz. – Kirby Lee Sep 02 '16 at 01:36
  • @KirbyLee Sure, this was an interesting problem. Note that I just corrected an important typo: I wrote $p(n+1)$ when I should have written $p(n+q)$. – Andrew Dudzik Sep 02 '16 at 13:01