Let $N\in\mathbb{N}$, and let $m,n$ be coprime. Also, suppose $a,b$ are relatively prime to $N$, and that $$ a^n\equiv b^n\mod{N},\ a^m\equiv b^m\mod{N} $$ I need to show that $a\equiv b\mod{N}$. I have an attempt, just one move I am not sure about.
Since $m,n$ are coprime, then for some $m',n'\in\mathbb{Z}$, $mm' + nn' = 1$. Now, since $a^n\equiv b^n\mod{N}$, then $a^{nn'}\equiv b^{nn'}\mod{N}$. From this, I get that $$ a^{nn'}\equiv b^{nn'}\mod{N} \Leftrightarrow a^{1-mm'}\equiv b^{1-mm'}\mod{N} $$
$$ \Leftrightarrow a(a^{-m})^{m'} \equiv b(b^{-m})^{m'}\mod{N} $$ Now, since $a,b$ are relatively prime to $N$, then $$ a^m\equiv b^m\mod{N} \Rightarrow a^{-m}\equiv b^{-m}\mod{N} $$ and therefore, again since $a,b$ are relatively prime to $N$, we can cancel them from both sides, leaving $$ a\equiv b\mod{N} $$
Are all the moves and derivations here correct?
One has $\exists u,v\in\Bbb{Z},\, um+vn=1$ as a direct consequence of $(m,n)=1$. Raising the first congruence to the power $v$ and the second to the power $u$ one gets $\pmod{N}$
$$\begin{align} a^{vn}&\equiv b^{vn}\\a^{um}&\equiv b^{um}\end{align}$$
Multiplying the two congruences together, one gets
$$a^{um+vn}=a\equiv b=b^{um+vn}\pmod{N}$$