Existence of Number to Power of Prime Minus 1 less 1 divisible by Prime Squared

Theorem

Let $p$ be a prime number.

Then there exists at least one positive integer $n$ greater than $1$ such that:

$n^{p - 1} \equiv 1 \pmod {p^2}$

Proof

\(\ds p^2\)

\(\equiv\)

\(\ds 0\)

\(\ds \pmod {p^2}\)

\(\ds 1\)

\(\equiv\)

\(\ds 1\)

\(\ds \pmod {p^2}\)

\(\ds \leadsto \ \ \)

\(\ds p^2 + 1\)

\(\equiv\)

\(\ds 0 + 1\)

\(\ds \pmod {p^2}\)

Modulo Addition is Well-Defined

\(\ds \)

\(\equiv\)

\(\ds 1\)

\(\ds \pmod {p^2}\)

\(\ds \leadsto \ \ \)

\(\ds \paren {p^2 + 1}^{p - 1}\)

\(\equiv\)

\(\ds 1^{p - 1}\)

\(\ds \pmod {p^2}\)

Congruence of Powers

\(\ds \)

\(\equiv\)

\(\ds 1\)

\(\ds \pmod {p^2}\)

Hence $p^2 + 1$ fulfils the conditions for the value of $n$ whose existence was required to be demonstrated.

$\blacksquare$

Examples

$p = 3$

The smallest positive integer $n$ greater than $1$ such that:

$n^{3 - 1} \equiv 1 \pmod {3^2}$

is $8$.

$p = 5$

The smallest positive integer $n$ greater than $1$ such that:

$n^{5 - 1} \equiv 1 \pmod {5^2}$

is $7$.

Sources

• 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $64$
• 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $64$