This beautiful problem was proposed by Mojtaba Zare.

War-Hammer wrote: This beautiful problem was proposed by Mojtaba Zare. but in my opinion it's absolutely very dirty and boring problem.

andria wrote: War-Hammer wrote: This beautiful problem was proposed by Mojtaba Zare. but in my opinion it's absolutely very dirty and boring problem. This problem has a 2 lined tricky solution and your opinion is not that much important

NDGA wrote: andria wrote: War-Hammer wrote: This beautiful problem was proposed by Mojtaba Zare. but in my opinion it's absolutely very dirty and boring problem. This problem has a 2 lined tricky solution and your opinion is not that much important First please read carefully i didn't write that the solution is long i told that the problem isn't nice and second i told it is my opinion maybe you love this problem...

It sufficient to prove that $p(x)$ is over $Q[X]$ We prove it by induction: Obviously leading coefficient is rational. we know that $a_n ,...a_{k+1}$ are rational. By cheking the coeficient of $[x^{n+k}]$ In equation we see that $a_k$ Is rational We are done.

Put the solution of this equation in problem: $5x^2+1=5x=j$ Then we have: $p(j)^2-3=p(j)$ $j$ is a number of the form $a+b\sqrt{5}$ And $p(j)$ is an number of the form$c+d\sqrt{13}$.but we know that coeficients Of $p(x)$ are rational. This is contradiction. Hence we're done!

Bump! I don't understand why we can assume that: $P(x)\in\mathbb{Q}[x]$

@above We didn't assume that $P(x)\in \mathbb{Q}[x]$, for non-constant case, we prove it by induction. Let $P(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$ where $n\geq 1,a_n\neq 0$. We prove by induction on $l\in \{ 1,2,...,n+1\}$ that $a_{n+1-l}\in \mathbb{Q}$. Comparing the coefficients of $x^{2n}$ gives us $a_n5^n=1\Rightarrow a_n=\frac{1}{5^n}\in \mathbb{Q}$. Now suppose $a_n,a_{n-1},...,a_{n+1-m}\in \mathbb{Q}$ for $m\in \{ 1,2,...,n\}$. We will show that $a_{n+1-(m+1)}\in \mathbb{Q}$. Comparing the coefficients of $x^{n+(n+1-(m+1))}$ gives us $5^{2n-m}\times \sum_{i=1}^{m+1}{a_{n+1-i}a_{n-(m+1)+i}}$ equal to $0$ (if $m$ is odd) or $\sum_{i =1}^{t+1}{a_{n+1-i}\binom{n+1-i}{n-t}5^{n-t}}$ (if $m$ is even and $t=\frac{m}{2}$). Note that $n-\frac{m}{2}\geq n+1-m$, so $RHS\in \mathbb{Q}$. This easily gives us $a_{n-m}\in \mathbb{Q}$, done.