Tomi's Challenges - #10

$F^9(F^{21}(c)-2021F^{20}(c))+F^3(F^{21}(c)-2021F^{20}(c))=F^9(4084441-2021F(c))+F^3(4084441-2021F(c))$

Let $a=F(F^{21}(c)-2021F^{20}(c))$ , $b=F(4084441-2021F(c))$ . We have:

$a^3-a=b^3-b$

Since $a^3-a$ is a bijection whose domain and range are both $\mathbb{R}$ , we can conclude that $a=b$ . We have:

$F(F^{21}(c)-2021F^{20}(c))=F(4084441-2021F(c))$

Since $F$ is strictly increasing, it's injective. We can conclude that this equation is therefore equivalent to:

$F^{21}(c)-2021F^{20}(c)=4084441-2021F(c)$

$F^{21}(c)-2021F^{20}(c)+2021F(c)-4084441=0$

The easiest (and possibly the only) way to find the sum of the solutions is to try to find them explicitly.

$F^{20}(c)(F(c)-2021)+2021(F(c)-2021)=0$

$(F(c)-2021)(F^{20}(c)+2021)=0$

From here we have $F(c)=2021$ , and the other factor yields complex solutions so we ignore it. From $F(20)=2021$ and the injectivity of the function, we conclude that $c=\boxed{20}$ .