7. $f(x)$ is a linear function, and the equation $f(f(x))=x+1$ has no solutions. Find all possible values of the quantity $f(f(f(f(f(2022)))))-f(f(f(2022)))-f(f(2022))$.
Solution. Let $f(x)=k x+b$, then $f(f(x))=k(k x+b)+b=k^{2} x+k b+b$. The equation can have no solutions only when $k^{2}=1$, that is, for functions $x+b$ or $-x+b$, so the answer is either $(2022+5 b)-(2022+3 b)-(2022+2 b)=-2022$, or $(-2022+b)-(-2022+b)-2022=-2022$.

Answer: -2022.

Criteria. Up to 2 points will be deducted for gaps in the proof that $k= \pm 1$.