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Parity of a Difference of Squares

Authors
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    Name
    Vu Hung
    Twitter

Problem Statement

Prove that if m,nm, n are integers, then m2−n2m^2 - n^2 is even if and only if at least one of (m+n)(m+n) and (m−n)(m-n) is even.

Hints

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

Solutions

Biconditional Proof

Step 1: Factor the expression

m2−n2=(m−n)(m+n)m^2 - n^2 = (m-n)(m+n)

This factorization will be used in both directions.

Forward Direction (  ⟹  \implies): If m2−n2m^2 - n^2 is even, then at least one of (m+n)(m+n) or (m−n)(m-n) is even.

Proof:

Since m2−n2m^2 - n^2 is even, the product (m−n)(m+n)(m-n)(m+n) is even.

A product of two integers is even if and only if at least one factor is even.

Therefore, at least one of (m−n)(m-n) or (m+n)(m+n) must be even. \blacksquare

Reverse Direction (  ⟸  \impliedby): If at least one of (m+n)(m+n) or (m−n)(m-n) is even, then m2−n2m^2 - n^2 is even.

Proof:

We consider two cases:

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

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

Then:

m2−n2=(m−n)(m+n)=(m−n)(2k)=2k(m−n)m^2 - n^2 = (m-n)(m+n) = (m-n)(2k) = 2k(m-n)

Since k(m−n)k(m-n) is an integer, m2−n2m^2 - n^2 is even.

Case 2: (m−n)(m-n) is even.

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

Then:

m2−n2=(m−n)(m+n)=(2j)(m+n)=2j(m+n)m^2 - n^2 = (m-n)(m+n) = (2j)(m+n) = 2j(m+n)

Since j(m+n)j(m+n) is an integer, m2−n2m^2 - n^2 is even.

In both cases, m2−n2m^2 - n^2 is even. \blacksquare

Conclusion:

Both directions proven, therefore:

m2−n2 is even  ⟺  at least one of (m+n),(m−n) is evenm^2 - n^2 \text{ is even} \iff \text{at least one of } (m+n), (m-n) \text{ is even}

Takeaways

  • Factorization First: Factoring m2−n2=(m−n)(m+n)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)(m+n) and (m−n)(m-n) always have same parity (both even or both odd), so "at least one even" is equivalent to "both even"

Further Readings

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

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