$$e(x) = (ax + b)$$ $$d(y) = a^{-1}(y-b)$$ I need to prove that $$e(x)=d(y)$$ iff $$(a^2)=1$$ and $$b(a+1)=0,$$ so I tried working with $d(ex)$ to show they can be equivalent:
$$ax + b = a^{-1}(ax + b - b).$$ Next step will be $$ax^2 + ba = ax$$ and that's where I'm stuck I saw a solution doing $$ ax+ b = a-1(x - b) $$ but I don't understand why this is valid. Shouldn't I replace the $y$ in $$d(y) = a^{-1}(y-b)$$ with the body of $$e(x)$$ after all $$x = d(e(x))$$ right?
And another question - how can I find all involutory keys in $\mathbb N$?