Question

Let $\sf p \geq 5$ be a prime number. Prove that there exists an integer $\sf a$ with $\sf 1 \leq a \leq p - 2$ such that neither $\sf {a}^{(p - 1)} - 1$ nor $\sf {(a + 1)}^{(p - 1)} - 1$ is divisible by $\sf p^2$​

Answer 1

Assume there exists two integers $p\geq5$ and $1\leq a\leq p-2$ such that $p^2\mid a^{p-1}-1$ and $p^2\mid (a+1)^{p+1}-1.$

We need to reach a statement that $p$ is not a prime number.

Let,

• $a^{p-1}=kp^2+1$

and,

• $(a+1)^{p-1}=jp^2+1$

Subtracting both,

$\longrightarrow(a+1)^{p-1}-a^{p-1}=(j-k)p^2$

$\displaystyle\longrightarrow\sum_{r=0}^{p-1}\,^{p-1}C_r\ a^{p-1-r}-a^{p-1}=(j-k)p^2$

$\displaystyle\longrightarrow a^{p-1}\left[\sum_{r=0}^{p-1}\,^{p-1}C_r\ a^{-r}-1\right]=(j-k)p^2$

$\displaystyle\longrightarrow a^{p-1}\left[1+\dfrac{p-1}{a}+\dfrac{(p-1)(p-2)}{2a^2}+\dots-1\right]=(j-k)p^2$

$\displaystyle\longrightarrow a^{p-1}(p-1)\left[\dfrac{1}{a}+\dfrac{p-2}{2a^2}+\dfrac{(p-2)(p-3)}{6a^3}+\dots\right]=(j-k)p^2$

Here each term in the series should be an integer.

Consider,

$\longrightarrow\dfrac{1}{a}\in\mathbb{Z}$

Since $a$ is also an integer,

$\Longrightarrow a=1$

Consider,

$\longrightarrow\dfrac{p-2}{2a^2}\in\mathbb{Z}$

$\Longrightarrow 2a^2\mid p-2$

$\longrightarrow p-2=2na^2\quad\quad\textrm{(Say)}$

$\longrightarrow p=2(na^2+1)$

This implies $p$ is not a prime number as prime numbers greater than 4 are indivisible by 2.

So we get,

$\begin{minipage}{11.5cm}\textrm{``If there exists two integers p\geq5 and 1\leq a\leq p-2 such that p^2\mid a^{p-1}-1 and p^2\mid (a+1)^{p-1}-1 , then p is not a prime number."}\end{minipage}$

Taking the contrapositive, we get,

$\begin{minipage}{11.5cm}\textrm{``If p\geq5 is a prime number, then there exists an integer 1\leq a\leq p-2 such that neither p^2\mid a^{p-1}-1 nor p^2\mid (a+1)^{p-1}-1. "}\end{minipage}$

Hence Proved!