Problem 22 · AMC 8 Stretch
Stretch
Number Theory
Geometry & Measurement
pattern-recognitionlogical-reasoning
Pick two whole numbers with \(0 < m < n\). Build \(p = n^2 - m^2\), \(q = 2mn\), \(r = m^2 + n^2\); these always satisfy \(p^2 + q^2 = r^2\). Using \(m = 1, n = 2\), find the value of \(p\) (the smaller leg).
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Answer: p = 3 (the triple is 3, 4, 5)
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Hint 1 of 4
First test the machine: with \(m = 1, n = 2\), compute \(p = n^2 - m^2\).
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Hint 2 of 4
\(p = 2^2 - 1^2\). Then \(q = 2 \times 1 \times 2\) and \(r = 1^2 + 2^2\). Do you recognize the triple?
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Hint 3 of 4
To see why it always works, compute \(p^2 + q^2 = (n^2 - m^2)^2 + (2mn)^2\) and simplify.
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Approach: Test the generator, then prove the identity by expanding
- With \(m = 1, n = 2\): \(p = 2^2 - 1^2 = 3\), \(q = 2 \times 1 \times 2 = 4\), \(r = 1^2 + 2^2 = 5\) — the famous \(3, 4, 5\) triple, and \(3^2 + 4^2 = 25 = 5^2\).
- Prove it always works: \((n^2 - m^2)^2 = n^4 - 2m^2 n^2 + m^4\) and \((2mn)^2 = 4m^2 n^2\).
- Adding, the middle terms combine: \(n^4 + 2m^2 n^2 + m^4 = (n^2 + m^2)^2 = r^2\), so \(p^2 + q^2 = r^2\) for every \(m < n\). (\(m=2, n=3\) gives \(5, 12, 13\).)
- So for \(m = 1, n = 2\) the smaller leg is \(p = 3\).
Mark:
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