Pinko wrote: Find all polynomials $P\in \mathbb{Q}[x]$, which satisfy the following equation: $P^2 (n)+\frac{1}{4}=P(n^2+\frac{1}{4})$ for $\forall$ $n\in \mathbb{N}$. From assertion, we get $P(x)^2=P(-x)^2$ and so, since polynomial, $P(x)$ is odd or $P(x)$ is even 1) The only odd solution is $P(x)\equiv x$ If $P(x)$ is odd : Let $P(x)=xQ(x^2)$ and equation becomes $x^2Q(x^2)^2+\frac 14=(x^2+\frac 14)Q((x^2+\frac 14)^2)$ And so, since polynomial, $xQ(x)^2+\frac 14=(x+\frac 14)Q((x+\frac 14)^2)$ If $Q(x)=1$ for some $x\ne -\frac 14$, above line implies $Q((x+\frac 14)^2)=1$ But setting two lines above $x=0$, we get $Q(\frac 1{16})=1$ And so $Q(a_n)=1$ for all elements of the sequence $a_1=\frac 1{16}$ and $a_{n+1}=(a_n+\frac 14)^2$ And since this sequence is infinite, the only possibility, since polynomial, is $Q(x)=1$ $\forall x$ And so $P(x)=x$ $\forall x$ Q.E.D. 2) If $P(x)$ is even Let $P(x)=Q(x^2)$ and equation becomes $Q(x^2)^2+\frac 14=Q((x^2+\frac 14)^2)$ And so, since polynomial : $Q(x)^2+\frac 14=Q((x+\frac 14)^2)$ Or also : $Q(x-\frac 14)^2+\frac 14=Q(x^2)$ Let then $R(x)=Q(x-\frac 14)$. Equation becomes $R(x)^2+\frac 14=R(x^2+\frac 14)$ So : if $P(x)$ is an even solution, then $\exists $ another solution $R(x)$ such that $P(x)=R(x^2+\frac 14)$ 3) Solution Combining both previous paragraphs, we easily get $\boxed{P(x)=f^{[n]}(x)}$ where $n\in\mathbb Z_{\ge 0}$ Where $f(x)=x^2+\frac 14$ and $f^{[n]}(x)$ is the composition of $f(x)$ $n$ times (with $f^{[0]}(x)\equiv x$)