I'm trying to understand the use of Euler's Totient Theorem in RSA. It is defined as : if $a$ is coprime with $n$ then $$ a^{\phi(n)}\equiv{1 modn} $$ I am unable to understand how the above implies that for any integers $m$ and $k$ the following statement holds : $$m^{k*mod\phi(n)}\equiv m^{k}mod n $$
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If $x=q\phi(n)+k$, then $m^x=m^{q\phi(n)+k}=(m^{\phi(n)})^qm^k\equiv1^qm^k=m^k$ – J. W. Tanner May 08 '20 at 12:58
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@J.W.Tanner for your proof you will also need the assumption that $\gcd(m,n)=1$, otherwise you will need CRT to make good of the claim. – Anurag A May 08 '20 at 13:00
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good point, @AnuragA – J. W. Tanner May 08 '20 at 13:15