# Finding all polynomials $P(x) \in \mathbb R[x]$ such that $P(x)^2=4P\left(x^2-5x+1\right)+2$

Find all polynomials $$P(x) \in \mathbb R[x]$$ such that $$P(x)^2=4P\left(x^2-5x+1\right)+2$$.

This comes from a no-solution class problem so it should have a definitive solution, unless my teacher wrongly wrote it as it is now.

Approach

This is like solving a functional equation, but only that it's a polynomial. I first thought I would solve for $$x=x^2-5x+1$$ or $$x^2-6x+1=0$$ which yields $$x_1=3+2\sqrt2$$ or $$x_2=3-2\sqrt2$$.

Which yields the value for $$P(x_1)$$ and $$P(x_2)$$. But then I couldn't proceed.

If $$d>0$$ and $$ax^d$$ is the leading term of $$P(x)$$, then $$a^2x^{2d}=4ax^{2d}$$, and thus $$a=4$$.

Another way I considered is using sequence to prove that there's infinitely many values of $$P(x)=P(y)$$ but that doesn't work either.

Any help is appreciated!

Let $$\alpha = 3 + 2 \sqrt 2$$, which satisfies $$\alpha ^ 2 - 6 \alpha + 1 = 0$$. Setting $$x = \alpha$$ in $$P ( x ) ^ 2 = 4 P \left ( x ^ 2 - 5 x + 1 \right ) + 2$$ we get $$P ( \alpha ) \in \left \{ 2 - \sqrt 6 , 2 + \sqrt 6 \right \}$$. By using strong induction on the positive integer $$n$$, we prove that $$P ^ { ( n ) } ( \alpha ) = 0$$, which shows that $$P$$ must be constant. Note that both such constant functions are indeed solutions. Assume that we know $$P ^ { ( m ) } ( \alpha ) = 0$$ for any positive integer $$m$$ less than $$n$$. Then we have $$\left ( \frac { \mathrm d } { \mathrm d x } \right ) ^ n \left ( P ( x ) ^ 2 \right ) = \left ( \frac { \mathrm d } { \mathrm d x } \right ) ^ n \Bigl ( 4 P \left ( x ^ 2 - 5 x + 1 \right ) + 2 \Bigr ) \text ;$$ $$\therefore \ 2 P ( x ) P ^ { ( n ) } ( x ) + S ( x ) = 4 ( 2 x - 5 ) ^ n P ^ { ( n ) } \left ( x ^ 2 - 5 x + 1 \right ) + T ( x ) \text .$$ Here $$S ( x )$$ is a sum each summand of which is the product of a term with $$P ^ { ( m ) } ( x )$$ for some positive integer $$m$$ less than $$n$$, and $$T ( x )$$ is a sum each summand of which is the product of a term with $$P ^ { ( m ) } \left ( x ^ 2 - 5 x + 1 \right )$$ for some positive integer $$m$$ less than $$n$$ (as the exact expressions for $$S ( x )$$ and $$T ( x )$$ are irrelevant to our proof, I summarized their description as such; but you can definitely do the tedious calculations and find the explicit formulas for them). By the induction hypothesis, we have $$S ( \alpha ) = T ( \alpha ) = 0$$, and hence $$\bigl ( P ( \alpha ) - 2 ( 2 \alpha - 5 ) ^ n \bigr ) P ^ { ( n ) } ( \alpha ) = 0 \text .$$ But note that $$( 2 \alpha - 5 ) ^ n = a + b \sqrt 2$$ for some integers $$a$$ and $$b$$ (the exact value of them is again irrelevant to the proof, but can be calculated explicitly in terms of $$n$$), which implies $$2 \pm \sqrt 6 \ne 2 ( 2 \alpha - 5 ) ^ n$$, and therefore we must have $$P ^ { ( n ) } ( \alpha ) = 0$$.