# Sum of Squares of Two Odd Integers is not Square

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## 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 $2.2$: $8 \ \text {(a)}$