# Problem Of The Week #390 Nov 1st, 2019

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#### anemone

##### MHB POTW Director
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Congratulations to the following members for their correct solution!

1. castor28
2. MegaMoh

Partial credit goes to lfdahl for his partially correct solution!

Solution from castor28 :
Let us write $f^n$ for the n-fold composition of $f$, so that $f^0(x) = x$ and $f^{n+1}(x) = f(f^n(x))$. We must solve $f^5(x)=0$.

The graph of $f$ has a minimum at $(-6,-6)$. This shows that $-6$ is a fixed point of $f$, and makes it worthwhile to try the substitution $x=z-6$. Under that substitution, we get:
$$f(z-6) = z^2 - 6$$
and, by induction,
$$f^n(z-6) = z^{2^n}-6$$
We must now solve:
$$f^5(x) = f^5(z-6) = z^{32}-6 = 0$$
giving the real roots $z=\pm\sqrt[32]{6}$ and $x=-6\pm\sqrt[32]{6}$.

The complex solutions are $x=-6 + \sqrt[32]{6}\,e^{\frac{2\pi ni}{32}}$, with $0\le n<32$.

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