As the title says, I am trying to prove that if the sum of the totatives of $a$ equals the sum of the totatives of $b$ then $a = b$ but I am stuck. I have that sum of totatives of $n = f(n)= \frac{n*\phi(n)}{2}$, but I am unsure of how to show that $f(a) = f(b) \implies a = b$.
I have the following:
$f(a) = \frac{a*\phi(a)}{2} = f(b) = \frac{b*\phi(b)}{2}$
$\implies a*\phi(a) = b*\phi(b)$
I am not sure where to go from here.
We are told that $a\varphi(a)=b\varphi(b)$, and want to conclude that $a=b$.
Suppose the result is true whenever $\max(a,b)\lt n$. We show the result is true when $\max(a,b)=n$.
Let $p$ be the largest prime that divides $a\varphi(a)$, or equivalently $b\varphi(b)$. Then $p$ is the largest prime that divides $a$, and $p$ is the largest prime that divides $b$. Moreover, $a=p^ka'$ and $b=p^kb'$ for some integer $k$, where neither $a'$ nor $b'$ is divisible by $p$. Thus $a\varphi(a)=p^ka'p^{k-1}(p-1)\varphi(a')$ and $b\varphi(b)=p^kb'p^{k-1}(p-1)\varphi(b')$.
Divide both $a\varphi(a)$ and $b\varphi(b)$ by $p^k\varphi(p^k)$. We get that $a'\varphi(a')=b'\varphi(b')$. By the induction hypothesis $a'=b'$ and therefore $a=b$.