Number which is Square and Cube Modulo 7/Proof 2

Theorem

Let $n \in \Z$ be an integer.

Let $n$ be both a square and a cube at the same time.

Then either:

$n \equiv 0 \pmod 7$

or:

$n \equiv 1 \pmod 7$

Proof

Let $n = r^2 = s^3$ for some $r, s \in \Z$.

Then:

$n = \paren {m^2}^3 = \paren {m^3}^2 = m^6$

for some $m \in \Z$

For $m \equiv 0 \pmod 7$:

\(\ds m\)

\(\equiv\)

\(\ds 0 \pmod 7\)

\(\ds \leadsto \ \ \)

\(\ds m^6\)

\(\equiv\)

\(\ds 0 \pmod 7\)

Congruence of Powers

\(\ds \leadsto \ \ \)

\(\ds n\)

\(\equiv\)

\(\ds 0 \pmod 7\)

From Fermat's Little Theorem, we have our prime number $p = 7$.

We also have our $m \in \Z_{>0}$ such that $7$ is not a divisor of $m$.

Then:

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

\(\equiv\)

\(\ds 1 \pmod p\)

\(\ds \leadsto \ \ \)

\(\ds m^{7 - 1}\)

\(\equiv\)

\(\ds 1 \pmod 7\)

\(\ds \leadsto \ \ \)

\(\ds m^6\)

\(\equiv\)

\(\ds 1 \pmod 7\)

\(\ds \leadsto \ \ \)

\(\ds n\)

\(\equiv\)

\(\ds 1 \pmod 7\)

Hence the result.

$\blacksquare$