Given that x is a real number, determine all polynomials with real coefficients satisfying this relationship:
P(x^{2}) = P(x)*P(x+2)
Among constant functions
P(x)=0 and P(x)=1 are clearly the only possibilities.
Among linear functions P(x)=ax+b
a little algebra shows a=1, b=1 is the only possibility.
So P(x)=x1.
Among quadratics P(x)=ax²+bx+c
a lot of algebra shows a=1, b=2, c=1 is the only possibility.
So P(x)=x²2x+1=(x1)²
From which appears a nice generalization:
P(x)=(x1)^n for all nonnegative integers n.
Proof:
(x^2 1)^n = (x1)^n*(x+1)^n
The
bold above are the solutions.
I did not check for other higher order polynomials. It seems these are the only ways to make the zeros to work out right.

Posted by Jer
on 20150627 16:40:37 