parmenides51 wrote: Find all polynomials $p(x)$ with real coefficients that satisfy $$4p(x^2) = 4(p(x))^2 + 4p(x)- 1$$ Constant solutions are $\boxed{\text{S1 : }P(x)=-\frac 12\quad\forall x}$ and $\boxed{\text{S2: }P(x)=\frac 12\quad\forall x}$ Nonconstant solutions with a unique summand are in the form $ax^n$ with $n\ge 1$ and $a\ne 0$ and, plugging in the equation, we easily see that no such solutions exist. For nonconstant solutions with at least two summands, let $ax^n+bx^p$ the two highest degree summands with $a,b\ne 0$ and $n>p\ge 0$ Looking at coefficient of $x^{2n}$ in both LHS,RHS, we get $4a=4a^2$ and so $a=1$ If $p\ge 1$, looking at coefficient of $x^{n+p}$ in both LHS,RHS, we get $0=8b$, impossible If $p=0$, then $P(x)=x^n+b$ and plugging this in equation, we get $b=-\frac 12$ And $\boxed{\text{S3 : }P(x)=x^n-\frac 12\quad\forall x}$, whatever is $n\ge 1$