Apply Fermat's little theorem to show that $A ^ { e d } \equiv A ( \bmod p )$

Problem : Let $$p$$, $$q$$ two distincts prime numbers and $$n=pq$$. Let $$e$$ be a prime number such that $$\gcd( (p-1)(q-1),e) = 1$$ and $$d$$ such that $$ed=1+k(p-1)(q-1)$$. Let $$A \in \mathbb { Z } / n$$. Show that $$A ^ { e d } \equiv A \pmod p$$.

My solution : We have that $$ed \equiv 1 (\bmod p-1)$$. Thus by Fermat's little theorem:

$$A ^ { e d } \equiv A^{k(p-1)+1} \equiv A^{k(p-1)}A^{1} \equiv ({A^{p-1}})^{k}A \stackrel{\text{Fermat}}{\equiv} 1^k A \equiv A \pmod p$$.

Question : Does the reasoning seems correct? Am I missing some steps?

• I would only suggest to choose another letter (not $k$) , but besides this : perfect solution. – Peter Apr 13 at 7:36
• @Peter Awesome, thank you very much! – NotAbelianGroup Apr 13 at 7:43