Sum of Squares of Two Odd Integers is not Square

Theorem

Let $m$ and $n$ be odd integers.

Then $m^2 + n^2$ is not a square number.

Proof

Aiming for a contradiction, suppose $m^2 + n^2$ is a square number.

Because $m$ and $n$ are both odd, we have:

\(\ds m^2\)

\(\equiv\)

\(\ds 1\)

\(\ds \pmod 4\)

Square Modulo 4

\(\ds n^2\)

\(\equiv\)

\(\ds 1\)

\(\ds \pmod 4\)

Square Modulo 4

\(\ds \leadsto \ \ \)

\(\ds m^2 + n^2\)

\(\equiv\)

\(\ds 2\)

\(\ds \pmod 4\)

From Parity of Integer equals Parity of its Square, $m^2$ and $n^2$ are both odd integers.

Hence $m^2 + n^2$ is an even integer.

But from Square Modulo 4, $m^2 + n^2$ is even if and only if $x^2 \equiv 0 \pmod 4$.

This contradicts the deduction that $m^2 + n^2 \equiv 2 \pmod 4$.

Hence by Proof by Contradiction it follows that $m^2 + n^2$ cannot be square.

$\blacksquare$

Sources

• 1980: David M. Burton: Elementary Number Theory (revised ed.) ... (previous) ... (next): Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Problems $2.2$: $8 \ \text {(a)}$