Is my proof that $f(x)$ where $f(f(x)) = 6x - f(x)$ for all $f:R+→R+$ is linear correct?

The top functional equation was assigned in a Putnam competition. To prove that this function is linear, I did the following algebra: $$f(f(x))=6x-f(x)$$ $$f(f(x)) + f(x) = 6x$$ $$\frac{f(f(x))+f(x)}{x}=6$$ Can I now say that, since this identity always results in a constant, $f(x)$ is a linear function?

Solution 1:

No, you cannot conclude this. From your remark in the comments, you recalled a video in which someone showed a claim like $$\require{cancel}g(x+1)-g(x)=2$$ and concluded that $g$ is linear. This sort of claim does not generalize to $$(\text{any sort of formula using $f$ twice})=(\text{const})\Rightarrow f\text{ is linear}$$

To see why the $g$-like claim works, suppose we can ensure that $x$ is an integer like $0,1,2,\dots$. Then $$g(3)=(g(3)-\cancel{g(2)})+(\cancel{g(2)}-\cancel{g(1)})+(\cancel{g(1)}-\cancel{g(0)})+g(0)=2+2+2+g(0)$$ In general, if $g(k)$ has $k$ summands, and so $g(k)=2k+g(0)$, which is linear. If $x$ is not an integer, then the claim is actually wrong: there are many solutions to the functional equation that are not linear.

In your particular example, there is no obvious way to iterate your construction to get many values of $x$. But note that when the functional equation was asked here, the accepted answer did develop a similar technique to solve it.