Parity Obstruction in Pythagorean Triples

Problem Statement

Prove that there is no Pythagorean triple $(a, b, c)$ where $a$ and $b$ (the two smallest numbers) are both even and $c$ (the largest number) is odd.

Hints

Use proof by contradiction. Assume such a triple exists with $a = 2k$ and $b = 2m$ (both even) and $c$ odd.

Substitute into the Pythagorean equation $a^2 + b^2 = c^2$ and analyze the parity of $c^2$ . What can you conclude about the parity of $c$ ?

Solutions

Proof by Contradiction:

Assume there exists a Pythagorean triple $(a, b, c)$ where:

$a^2 + b^2 = c^2$
$a$ and $b$ are both even
$c$ is odd

Since $a$ and $b$ are even, write $a = 2k$ and $b = 2m$ for integers $k, m$ .

Substitute into Pythagorean equation:

\begin{aligned} (2k)^2 + (2m)^2 &= c^2 \\ 4k^2 + 4m^2 &= c^2 \\ 4(k^2 + m^2) &= c^2 \end{aligned}

Analyze parity of $c^2$ :

The equation shows $c^2 = 4(k^2 + m^2)$ , which is a multiple of $4$ .

In particular, $c^2$ is divisible by $2$ , so $c^2$ is even.

Derive parity of $c$ :

If $c$ were odd, then $c = 2n+1$ for some integer $n$ , and:

c^2 = (2n+1)^2 = 4n^2 + 4n + 1 = 2(2n^2 + 2n) + 1

This shows $c^2$ would be odd, contradicting that $c^2$ is even.

Therefore, $c$ must be even.

Contradiction:

We derived that $c$ must be even, which contradicts our assumption that $c$ is odd.

Hence, no such Pythagorean triple exists. $\blacksquare$

Takeaways

Reconstruct the full proof from the hint and the solution outline, and justify every transformation explicitly.
Check edge cases and verify where each assumption is used in the argument.

Further Readings

If you found this proof interesting, be sure to check out these relevant HSC booklets to sharpen your reasoning skills:

HSC Proofs: https://vumaths.com/booklets/hsc-proofs/
HSC Distributions: https://vumaths.com/booklets/hsc-distributions/
HSC Last Resorts: https://vumaths.com/booklets/hsc-last-resorts/

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