Question

Let P(x) be a polynomial of degree 2m.
Show that \( P(x^2)= P(x)P(x-1) \) if and only if
\[ P(x) =\left( x^2 + x+1 \right)^m\]

Answer 1

See
http://openstudy.com/study#/updates/4ff6b092e4b01c7be8c98dee
for the particular case where m=5.

Answer 2

hmm looks like all the heavy lifting was done showing that since the only roots are \(-\frac{1}{2}\pm\frac{\sqrt{3}}{2}i\) it must be \((x^2+x+1)^m\)
is that the hook, or is there something else?

Answer 3

I agree. I posted so people can see that it is true in this general form.

Answer 4

ok then kudos to @mukushla who did all the work for this one