It is not true, because there are infinetely many solutions $y^{2}-2x^{2}=1$. It mean $P(x)=2x^{2}+1=y^{2}$ for infinetely many x.

That is not a monic polynomial.

Yes. I forgot about monic. If $P(x)=x^{2n}+a_{2n-1}x^{2n-1}+...+a_{0}$, then $4P(x)=Q(x)^{2}+R(x), \ Q(x)=2x^{n}+b_{n-1}x^{n-1}+...+b_{0})^{2}, \ R(x),$ were $deg(R(x))\le n-1.$ Therefore for big x $|R(x)|<2|Q(x)|-1$. Because $4P(x)$ and $Q(x)^{2}$ are perfect square, it mean $R(x)=0$ for infinetely many x, or $R(x)\equiv 0.$ Because all coefficients $Q(x)^{2}$ divide by 4, by induction we get all coefficients Q(x) are even.

These [equations of Pell type] are, essentially, the only exceptions. A special case of Siegel's theorem on integral points on algebraic curves ("there are at most finitely many integral points on an affine algebraic curve $C$ with at least three points at infinity, i.e. such that $\# \tilde{C}\setminus C \geq 3$, where $\tilde{C}$ is the projective closure of $C$ [relative to the considered embedding in affine space]") implies: if $P(x) \in \mathbb{Z}[x]$ is an integer polynomial with at least three distinct roots, then the equation $y^{2}= P(x)$ has only finitely many solutions in integers $x,y$. In particular, any irreducible polynomial of degree $\geq 3$ takes at most finitely many perfect square values. A proof of this is possible via Roth's theorem of diophantine approximation. (And, as well, other results of diophantine approximations, such as the $2$-variable $S$-unit equation, which can be solved effectively). The general version of Siegel's theorem, valid for arbitrary curves, is proven standardly by embedding a curve in its Jacobian variety and then using Roth's theorem and the arithmetic of abelian varieties. But more elegant (and short, granted "the subspace theorem" of W. Schmidt, which is a far-reaching extension of Roth's theorem) is a relatively recent proof, by Corvaja and Zannier, which doesn't at all use the arithmetic of abelian varieties! But of course, for the special case you have proposed, no algebraic geometry is needed; it suffices to know what is the "2-variable $S$-unit equation" for number fields ("if $K$ is a number field and $S \subset M_{K}$ any finite set of places, then the equation $x+y = 1$ has only finitely many solutions in $S$-units, i.e. with $|x|_{v}, |y|_{v}\leq 1$ for all $v \notin S$"); it is then straightforward to deduce that $y^{2}= P(x)$ has only finitely many solutions, provided that $P(\cdot)$ has at least three complex roots. It is a much deeper theorem that if $P(x)$ is an irreducible polynomial of degree at least $5$, then $y^{2}= P(x)$ has at most finitely many solutions in rational numbers $x,y$; this was proven only in 1983 by G. Faltings (other proofs are known since then, but they are also extremely deep). For the problem that you propose, you want the degree to be even and $>2$ in order to bound a square between consecutive squares.

Rust wrote: Because all coefficients $ Q(x)^{2}$ divide by 4, by induction we get all coefficients Q(x) are even. Can you show this induction?

Rust wrote: then $ 4P(x) = Q(x)^{2} + R(x), \ Q(x) = 2x^{n} + b_{n - 1}x^{n - 1} + ... + b_{0})^{2}, \ R(x),$ were $ deg(R(x))\le n - 1.$ Actually I don't agree with this! Considering the degree $ x_{2n-1}$, you get $ 4b_{n-1}=4a_{2n-1}$, so $ b_{n-1}=a_{2n-1}$. Considering the degree $ x_{2n-2}$, you get $ 4b_{n-2}+b_{n-1}^2=4a_{2n-2}$, so if $ b_{n-1}=a_{2n-1}$ is odd, you cannot choose $ b_{n-2}$ to be an integer...

Let me certify Rust's correct argument. Assume $ P(x)\in\mathbb{Z}[x]$ with degree $ 2n$ and leading coefficient $ c^2$ for some $ n,c\in\mathbb{N}$, and $ P(m)$ is a perfect square for infinitely many $ m\in\mathbb{Z}$. Then $ p(x) = x^{2n}P({1\over x})\in\mathbb{Z}[x]$ of degree $ \le 2n$ and $ p(0) = c^2$. So we can write $ p(x) = c^2 + xt(x)$ for some $ t(x)\in\mathbb{Z}[x]$. Then using Newton's generalized binomial theorem we get $ p(x) = \left(\sum_{k\ge 0}\binom{1/2}{k}c^{1 - 2k}x^kt(x)^k\right)^2$ with $ \sum_{k\ge 0}\binom{1/2}{k}c^{1 - 2k}x^kt(x)^k\in\mathbb{Q}[ [x] ]$. So $ p(x)\equiv q(x)^2\pmod{x^{n + 1}}$ for some $ q(x)\in\mathbb{Q}[x]$ of degree $ \le n$. Then $ p(x) = q(x)^2 + x^{n + 1}r(x)$ for some $ r(x)\in\mathbb{Q}[x]$, and $ x^{n + 1}r(x) = p(x) - q(x)^2$ has degree $ \le 2n$, so $ \deg r(x)\le n - 1$. And now $ P(x) = x^{2n}p({1\over x}) = x^{2n}(q({1\over x})^2 + {1\over x^{n + 1}}r({1\over x})) = Q(x)^2 + R(x)$ with $ Q(x): = x^nq({1\over x})\in\mathbb{Q}[x]$ of degree $ \le n$ and $ R(x): = x^{n - 1}r({1\over x})\in\mathbb{Q}[x]$ of degree $ \le n - 1$. Let $ d\in\mathbb{N}$ be minimal with $ dQ(x)\in\mathbb{Z}[x]$. Then $ \deg P(x) = 2n$ gives $ \deg Q(x) = n > \deg R(x)$, and therefore $ 2|dQ(x)| - 1 > |d^2R(x)|$ for sufficiently large $ |x|$. Then $ d^2P(x) = |dQ(x)|^2 + d^2R(x) > |dQ(x)|^2 - 2|dQ(x)| + 1 = (|dQ(x)| - 1)^2$ and $ d^2P(x) = |dQ(x)|^2 + d^2R(x) < |dQ(x)|^2 + 2|dQ(x)| - 1 < (|dQ(x)| + 1)^2$ for sufficiently large $ |x|$. So if $ P(m)$ is a perfect square for sufficiently large $ |m|$ with $ m\in\mathbb{Z}$, then so is $ d^2P(m)$, which gives $ d^2P(m) = (|dQ(m)|)^2$ and $ P(m) = Q(m)^2$. But there are infinitely many such $ m$, which gives $ P(x) = Q(x)^2$. And now if $ d > 1$ we take $ p\mid d$ prime, and then $ (dQ(x))^2 = d^2P(x)\equiv 0 \pmod{p}$. But $ \mathbb{F}_p = \mathbb{Z}/p\mathbb{Z}$ is a field, so $ \mathbb{F}_{p}[x]$ is an Euclidean domain, and therefore has no zero divisors. This gives $ dQ(x)\equiv 0 \pmod{p}$ and $ {d\over p}Q(x)\in\mathbb{Z}[x]$, which contradicts the minimal choice of $ d$. So $ d = 1$ and $ Q(x)\in\mathbb{Z}[x]$.

We can generalize this problem: Let $P(x)\in \mathbb{Z}[x]$ be a monic polynomial with degree divisible by an integer $d$. Prove that, if for infinitely many integers $x$, the number $P(x)$ is a $d$th power of a positive integer, then there exists a polynomial $Q(x)\in\mathbb{Z}[x]$ such that $P(x)=Q(x)^d$.

bilarev wrote: Let $P(x)\in \mathbb{Z}[x]$ be a monic polynomial with even degree. Prove that, if for infinitely many integers $x$, the number $P(x)$ is a square of a positive integer, then there exists a polynomial $Q(x)\in\mathbb{Z}[x]$ such that $P(x)=Q(x)^2$. It would be a nice problem: Prove that if $P(x)\in Z[x]$ is a perfect square for all $n\in N$ then there exists $Q(x)\in Z[x]$ such that $P(x)=Q(x)^2$ Use this problem to prove the above problem

K.I.M wrote: bilarev wrote: Let $P(x)\in \mathbb{Z}[x]$ be a monic polynomial with even degree. Prove that, if for infinitely many integers $x$, the number $P(x)$ is a square of a positive integer, then there exists a polynomial $Q(x)\in\mathbb{Z}[x]$ such that $P(x)=Q(x)^2$. It would be a nice problem: Prove that if $P(x)\in Z[x]$ is a perfect square for all $n\in N$ then there exists $Q(x)\in Z[x]$ such that $P(x)=Q(x)^2$ Use this problem to prove the above problem well in that case we need to prove the following statement for $\exists n \in \mathbb{N}, \exists x \in \mathbb{Z}$, $P(x)=n^2$ then prove that $\forall x \in \mathbb{Z}$, $P(x)=n^2$ for proper $n$s

Skravin wrote: K.I.M wrote: bilarev wrote: Let $P(x)\in \mathbb{Z}[x]$ be a monic polynomial with even degree. Prove that, if for infinitely many integers $x$, the number $P(x)$ is a square of a positive integer, then there exists a polynomial $Q(x)\in\mathbb{Z}[x]$ such that $P(x)=Q(x)^2$. It would be a nice problem: Prove that if $P(x)\in Z[x]$ is a perfect square for all $n\in N$ then there exists $Q(x)\in Z[x]$ such that $P(x)=Q(x)^2$ Use this problem to prove the above problem well in that case we need to prove the following statement for $\exists n \in \mathbb{N}, \exists x \in \mathbb{Z}$, $P(x)=n^2$ then prove that $\forall x \in \mathbb{Z}$, $P(x)=n^2$ for proper $n$s Could you please explain more? I don't get what you say

K.I.M wrote: Could you please explain more? I don't get what you say I used some letters to lessen the statement but it had rather distracted "For infinitely many(not could be all) positive integers, $P(x)$ becomes the square number. Then prove that $P(x)$ becomes the square number for all $x \in \mathbb{Z^+}$

i dont think its this hard. actually we need to prove that there exists a an positive integer m such that (x+m)^2 < P(x) < (x+m+1)^2 ( for all x>N0) . ofcourse if we assume that p(x) is not Q(x)^2 . and if we assume that there exists infinitely many POSITIVE integer r such that p(r) is square then things will be trivial . if not then there exists infinitely many negative integers and so we may need to prove there is a m which (x-m)^2 < p(x) < (x-m+1)^2 . which is not hard to prove . its the summary of my solution . but i can write it completely . just inform me

olorin wrote: [...] And now if $ d > 1$ we take $ p\mid d$ prime, and then $ (dQ(x))^2 = d^2P(x)\equiv 0 \pmod{p}$. But $ \mathbb{F}_p = \mathbb{Z}/p\mathbb{Z}$ is a field, so $ \mathbb{F}_{p}[x]$ is an Euclidean domain, and therefore has no zero divisors. This gives $ dQ(x)\equiv 0 \pmod{p}$ and $ {d\over p}Q(x)\in\mathbb{Z}[x]$, which contradicts the minimal choice of $ d$. So $ d = 1$ and $ Q(x)\in\mathbb{Z}[x]$. Can anybody deal with this part without using field or domain, i.e in more elementary way? Or you have another more simple solution? Thanks very much!