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Let $f(x)\in \mathbb{Z}[x] - \mathbb{Z}$ with all roots distinct in $\mathbb{C}$. I need to show that for infinitely many $n\in\mathbb{Z}$ we have $f(n)$ is not a perfect square. Any help would be greatly appreciated :)

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    $\begingroup$ I don't know if this will work, but if $r$ is a quadratic residue mod $m$ and you find one solution to $f(n) \equiv r \pmod{m}$ then all $r+km$ satisfy your condition. $\endgroup$ – B. Goddard Nov 29 '19 at 11:51
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Consider the polynomial $F(x,y)=y^2-f(x)$. It is obviously irreducible since $f$ is square-free in $\mathbb{Q}[x,y]$ and thus we can apply Hilbert's Irreducibility Theorem to obtain an infinitude of $x_i$ such that $y^2-f(x_i)=0$ is irreducible or equivalently $f(x_i)\ne{}n^2,\forall{}n\in\mathbb{N}$. This means that the infinite $f(x_i)$ we found are not perfect squares.

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  • $\begingroup$ Do you know a more elementary proof by any chance? $\endgroup$ – choroids gaga Nov 29 '19 at 13:00
  • $\begingroup$ Sadly nothing comes to mind $\endgroup$ – Μάρκος Καραμέρης Nov 29 '19 at 15:58
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According to this https://mathoverflow.net/q/224881 it is a "well-known" fact that if $f(n)$ is a perfect square for every integer $n$, then $f(x)$ is a perfect square (as a polynomial.) Since the roots of your polynomial are distinct, it can't be a perfect square, so there exists an integer $m$ such that $f(m) = b$ is not a perfect square.

Then, according to this https://math.stackexchange.com/a/646135/362009 there exists a prime $p$ for which $b$ is a quadratic non-residue. Then for any integer $k$ we have $f(m+kp) \equiv f(m) \equiv b \pmod{p}$ and so we have infinitely many integers for which $f(n)$ can't be a square.

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