Parity Obstruction in Pythagorean Triples

Parity (odd/even) arguments are a fundamental tool in the HSC Proofs topic, appearing from Extension 1 through to Extension 2. In this problem we use a contradiction argument to show that a Pythagorean triple cannot have both smaller legs even while the hypotenuse is odd. By carefully tracking the parity of squares, you’ll see how a simple algebraic substitution can quickly expose an inconsistency — a pattern you can apply to many other number-theoretic proofs.

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.

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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 $c$?

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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 $k, m$. We can now substitute these expressions into the Pythagorean equation to isolate a factor of $4$ on the left‑hand side.

Substitute into Pythagorean equation:

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

The resulting equation shows $c^2$ is a multiple of $4$, so $c^2$ is certainly even. This is the crucial observation that lets us test the parity of $c$ itself.

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.

But if $c$ itself is odd, what would its square be? Let’s check directly.

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.

We have arrived at a contradiction: we started by assuming $c$ is odd, but the algebra forced $c$ to be even. No such Pythagorean triple can exist.

Contradiction:

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

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

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Takeaways

• Proof by contradiction often succeeds when a simple invariant such as parity can force an inconsistency from the assumed conditions.
• 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.

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Further Readings

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