Clearly F is a quadratic with leading coefficient=1
F(x)=(x-p)(x-q)
And the equation becomes
((x-p)(x-q)-p)((x-p)(x-q)-q) = (x-p)(x-q)(x^2+x+1)
(x^2-(p+q)x+pq-p)(x^2-(p+q)x+pq-q) = (x^2-(p+q)x+pq)(x^2+x+1)
If the second factor of each pair are equal then
-(p+q)=1 and pq-q=1
p+q+1=0 and p=1+1/q
1+1/q+q+1=0
q^2+2q+1=0
q=-1 and p=0
Switching to pair the first of the LHS with the second of the RHS just switches p and q.
F(x) = x(x+1) = x^2+x
Checking F(F(x)) = (x^2+x)^2+(x^2+x)
= x^4 + 2x^3 + 2x^2 + x
= (x^2+x)(x^2+x+1)
= F(x)(x^2+x+1)
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Posted by Jer
on 2024-11-27 14:14:39 |
solution
