If $gcd(n,m)=1$ and $x\equiv y(mod\phi(m))$ then $n^{x}\equiv n^{y}(modm)$.

Question

I am trying to prove that if $gcd(n,m)=1$ and $x\equiv y(mod\phi(m))$ then $n^{x}\equiv n^{y}(modm)$. Any help would be much appreciated!

By Euler's theorem, if $\gcd(n,m)=1$ then $n^{\phi(m)}\equiv1\pmod m$ — J. W. Tanner, Nov 07 '19 at 02:12
See the Theorem & Corollary in the linked dupe (using $,e=\phi(m)),$ & Euler's Theorem) — Bill Dubuque, Nov 07 '19 at 02:22

score -1 · Accepted Answer · answered Nov 07 '19 at 02:13

-1

$\text{gcd}(n,m)=1\implies n^{\phi(m)}\equiv 1(\text{ mod } m)$.

Now, $x\equiv y(\text{ mod }\phi(m))\implies x=y+k\phi(m)$ for some integer $k$.

Now, $$n^x=n^{y+k\phi(m)}=n^y\cdot \big(n^{\phi(m)}\big)^k\equiv n^y\cdot (1)^k=n^y(\text{ mod } m).$$

answered Nov 07 '19 at 02:13

Sumanta

1 Answers1