Parity of a Difference of Squares

This problem explores the parity of a difference of squares, linking algebraic factorisation to number theory. It's a classic exercise in biconditional proof, directly relevant to the HSC Mathematics Extension 1 Proofs topic. By working through it, you'll strengthen your ability to use factorisation and parity properties (even/odd product and sum) in logical arguments.

Problem Statement

Prove that if $m, n$ are integers, then $m^2 - n^2$ is even if and only if at least one of $(m+n)$ and $(m-n)$ is even.

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Hints

Attempt the proof independently first. Focus on the key theorem, algebraic transformation, or contradiction setup that links the hypothesis to the target conclusion.

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Solutions

Biconditional Proof

Step 1: Factor the expression

$$m^2 - n^2 = (m-n)(m+n)$$

This factorization will be used in both directions.

Forward Direction ($\implies$): If $m^2 - n^2$ is even, then at least one of $(m+n)$ or $(m-n)$ is even.

Proof:

We prove the forward implication. Assuming $m^2 - n^2$ is even, the product $(m-n)(m+n)$ is even. A fundamental parity rule says: a product of two integers is even if and only if at least one factor is even. Therefore, at least one of $(m-n)$ or $(m+n)$ must be even.

Reverse Direction ($\impliedby$): If at least one of $(m+n)$ or $(m-n)$ is even, then $m^2 - n^2$ is even.

Proof:

Now we assume at least one of the factors is even, and we prove that $m^2 - n^2$ is even. We consider two cases:

Case 1: $(m+n)$ is even.

Write $(m+n) = 2k$ for some integer $k$.

Then:

$$m^2 - n^2 = (m-n)(m+n) = (m-n)(2k) = 2k(m-n)$$

Since $k(m-n)$ is an integer, $m^2 - n^2$ is even.

Case 2: $(m-n)$ is even.

Write $(m-n) = 2j$ for some integer $j$.

Then:

$$m^2 - n^2 = (m-n)(m+n) = (2j)(m+n) = 2j(m+n)$$

Since $j(m+n)$ is an integer, $m^2 - n^2$ is even.

In both cases, $m^2 - n^2$ is even. \blacksquare

Conclusion:

Both directions proven, therefore:

$$m^2 - n^2 \text{ is even} \iff \text{at least one of } (m+n), (m-n) \text{ is even}$$

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Takeaways

• Factorization First: Factoring $m^2 - n^2 = (m-n)(m+n)$ immediately connects to parity of factors
• Product Parity: Product even $\iff$ at least one factor even (fundamental parity property)
• Case Analysis: Reverse direction handles two cases (either factor even) separately
• Note on Parity: Actually, $(m+n)$ and $(m-n)$ always have same parity (both even or both odd), so "at least one even" is equivalent to "both even"

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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 Vectors, HSC Trigonometry