Question

Prove Fermat’s little theorem, which is given in Theorem 10.6. (Hint: Consider the sequence a1, a2, . . . What must happen, and how?)

THEOREM 10.6.

If p is prime and,$\mathbf{a}\in {{\mathbf{{\rm Z}}}_p}^{+}$then.

Short answer

Using the definition of Fermat’s little Theorem which states that if$p$ is a prime number, then for any integer $a$, the number$a^p-a$ is an integer multiple of,$p$ the above problem is solved.

Step-by-step solution

Understanding the theorem 

Statement: if$p$be a prime and then.$a^{p-1}\equiv 1\left(\mathrm{mod}p\right)$ Here${{{\rm Z}}_p}^{+}$ is defined as ${{{\rm Z}}_p}^{+}=\{1,\dots ,p-1\}$and$(p,a)$ is co-prime. It is also referred to as a miniature Fermat's theorem and is occasionally written as .$a^{p-1}\equiv 1\left(\mathrm{mod}p\right)$

Take into account the first$p-1$ positive multiple of.$a$

$a,2a,3a,\dots ,(p-1)a$

As the little Fermat’s theorem states$(p,a)$“is co-prime (that is, $p$is not exactly divisible by$a$)”. Suppose $xa$and ya are taken in such a way that, the modulo p of xa and ya are equal.

Continuing with the proof of the theorem 

Now, it can be said that .$x=s\left(\mathrm{mod}p\right)$ So, $p-1$the multiples by a above are non-zero and distinct; that is, they must be congruent to $a,2a,3a,\dots ,(p-1)a$in the same order. Now, multiply all these congruence together and which gives:

$a,2a,3a,\dots ,(p-1)a=1\times 2\times 3\times \dots \times (p-1)\left(\mathrm{mod}p\right)$

$a^{n-1}(p-1)!=(p-1)!\left(\mathrm{mod}p\right)$

Now, dividing each side by $(p-1)!$in the above equality.$a^{p-1}\equiv 1\left(\mathrm{mod}p\right)$