Question

It has been over 1 hour, so here is the proof:

Answer 1

Find all Polynomials such that:\[f(x^2) + f(x)f(x+1) = 0\]

Answer 2

therefore:
\[f(x)f(x+1) = -f(x^2)\]

Answer 3

substitute in:
\[g(x) = -f(x)\]

Answer 4

\[g(x)g(x+1) = g(x^2)\]

Answer 5

now, we know that if:
\[g(a) = 0 \]
\[g(a^2) = 0\]

Answer 6

therefore, if the modulus of a is greater than one, or between one and 0, we will have an infinite number of roots

Answer 7

now, we write the relationship:
\[g(x) = \frac{g(x^2)}{g(x+1)} = \frac{g((x-1)^2)}{g(x-1)}\]

Answer 8

any questions so far?

Answer 9

so, if a is a root, (a-1)^2 is a root

Answer 10

therefore:
\[|(a-1)^2| = 1,0\]

Answer 11

therefore:
\[|a-1|^2 = 1,0\]

Answer 12

therefore:
\[|a-1| = 1,0\]

Answer 13

so, there are only four numbers such that:
\[|a| = |a-1| = 1,0\]
1,0,e^(i*pi/3), and e^(i*5*pi/3)