IRL_ABooklet

The two solutions are $f(x)=x$ and $f(x)=-x$. We prove this in three stages. First we show that $f$ is self-inverse, that is, $f(f(x))=x$ for all integers $x$. Secondly we show that $f$ is additive. Thirdly, we demonstrate that the stated solutions are the only self-inverse additive functions.

To show the self-inverse property, interchange $a$ and $b$ in the original equation $$f(b+f(a))=a+f(b),$$ then apply $f$ again to each side, which gives $$f(f(b+f(a)))=f(a+f(b))=b+f(a).$$ Any integer can be represented as $b+f(a)$ hence $f(f(x))=x$ for all $x$. For additivity, let $c=f(b)$. By the self-inverse property, any integer $c$ can be written in this form, setting $b=f(c)$. The original equation becomes $$f(a+c)=f(c)+f(a).$$ Putting $a=c=0$ implies $f(0)=0$. It follows by induction that $f(x)=xf(1)$ for positive integers $x$. Writing $c=-a$ proves that $f(x)=xf(1)$ for all negative $x$. Therefore, $f(x)$ is linear with slope $f(1)$ and $y$-intercept of zero. Finally, the self-inverse property with $x=1$ gives $1=f(f(1))=f(1{)}^2$, whence $f(1{)}^2=1$ and $f(1)=\pm 1$ yielding the two solutions claimed. It is straightforward to check that both indeed are solutions.