deletions | additions       diff --git a/section_On_Fermat_s_Little__.tex b/section_On_Fermat_s_Little__.tex  new file mode 100644  index 0000000..fe7245f  --- /dev/null  +++ b/section_On_Fermat_s_Little__.tex 
...
\section*{On Fermat's Little Theorem}  We said that if we take a prime number $p$, $\forall a \in \mathbb{F}_{p}^{*}$, the Fermat's Little Theorem says us   $$a^{p-1} \equiv 1 \ \ mod \ p $$  It's easy to verify that if we take two primes $p$ and $q$, then  $$a^{(p-1)(q-1)} \equiv 1 \ \ mod \ pq $$  because   $$a^{(p-1)(q-1)} \ mod \ pq \ \ \Rightarrow \begin{cases}  a^{(p-1)(q-1)} \ \ mod \ p \ \ \Rightarrow (a^{p-1})^{q-1} \equiv_{p} 1 \\   a^{(p-1)(q-1)} \ \ mod \ q \ \ \Rightarrow (a^{q-1})^{p-1} \equiv_{q} 1   \end{cases}$$  Then we could present this preposition:  \begin{pps}  Let $p,q$ be primes, $N=p \cdot q$ and $e \geq 1$ such that $gcd(e,(p-1)(q-1))$; we consider $d$ such that   $$ed \equiv 1 \ mod \ (p-1)(q-1) $$  $$THEN$$  the equation $$x^{e} \equiv c \ mod \ N \ \ \ c \in \mathbb{Z}_{N}$$  admitts a unique solution   $$x \equiv c^{d} \ mod \ N $$  \end{pps}  \begin{proof}  We could quickly observe that considering  $$x^{e} \equiv c \ mod \ N$$  and we power it by $d$  $$x^{ed} \equiv c^{d} \ mod \ N $$  then we use the fact that   $$ed \equiv 1 \ mod \ (p-1)(q-1) \ \ \Rightarrow ed= 1 + k(p-1)(p-1)$$  and substituting in the equation above we get:  $$x^{ed} \equiv_{N} x^{1 + k(p-1)(p-1)} \equiv_{N} x \cdot (x^{(p-1)(q-1)})^{k} \equiv_{N} x $$  \end{proof}  This result will be important to set up the RSA Cryptosystem.