Answer to Question #110947 in Algebra for Madhav

Question

Let Px be a quadratic polynomial such that for distinct realms alpha and beta p-alpha =alpha p-beta=beta show that alpha and beta are roots of p(p(x)) -x=0

Answer 1

Let $P(x)$ be a quadratic polynomial such that for distinct realms $\alpha$ and $\beta$ $P(-\alpha) =\alpha, P(-\beta)=\beta$ show that alpha and beta are roots of $P(P(x)) -x=0$

Let $P(x)=a_0x^2+a_1x+a_2, a_0,a_1,a_2\in R, a_0\neq0$

$P(-\alpha)=a_0\alpha^2-a_1\alpha+a_2=\alpha\\ a_0\alpha^2+a_2=\alpha+a_1\alpha$

$P(P(x))-x=a_0(a_0x^2+a_1x+a_2)^2+\\ +a_1(a_0x^2+a_1x+a_2)+a_2-x$

Then

$P(P(\alpha))-\alpha=a_0(a_0\alpha^2+a_1\alpha+a_2)^2+\\ +a_1(a_0\alpha^2+a_1\alpha+a_2)+a_2-\alpha=\\ =a_0(\alpha+2a_1\alpha)^2+a_1(\alpha+2a_1\alpha)+a_2-\alpha=\\ =a_0\alpha^2+4a_0a_1\alpha^2+4a_0a_1^2\alpha^2+a_1\alpha+2a_1^2\alpha+\\ +a_2-\alpha=\alpha+a_1\alpha+4a_0a_1\alpha^2+4a_0a_1^2\alpha^2+\\ +a_1\alpha+2a_1^2\alpha-\alpha=\\ =4a_0a_1\alpha^2+4a_0a_1^2\alpha^2+\\ +2a_1\alpha+2a_1^2\alpha=4a_0a_1\alpha^2(1+a_1)+\\ +2a_1\alpha(1+a_1)=\\=2a_1\alpha(1+a_1)(2a_0\alpha+1)=0$

becase $a_1=-1$

$P(1)=a_0+a_1+a_2$ - the sum of the coefficients of the polynomial

$a_0-a_1+a_2-1=0$

Similarly $P(P(\beta))-\beta=0$

or

$P(P(\beta))-\beta=P(-(-P(\beta)))-\beta=-P(\beta)-\beta=\\ -P(-(-\beta))-\beta=-(-\beta)-\beta=\beta-\beta=0$