Which pair (m, n) satisfies ALL the conditions required to generate a primitive Pythagorean triple via a = m² − n², b = 2mn, c = m² + n²?
Am = 6, n = 2 — m > n, but gcd(6, 2) = 2 and both are even
Bm = 3, n = 1 — m > n and gcd = 1, but both are odd (same parity)
Cm = 5, n = 3 — m > n and gcd = 1, but both are odd (same parity)
Dm = 5, n = 2 — m > n, gcd(5, 2) = 1, and opposite parity
The conditions are: m > n > 0, gcd(m, n) = 1, and m and n have opposite parity (one odd, one even). Options A fails coprimality (gcd = 2). Options B and C fail the parity condition — both m and n are odd. Option D: m = 5 (odd), n = 2 (even), gcd(5, 2) = 1, m > n. This satisfies all conditions. The resulting triple is a = 25 − 4 = 21, b = 20, c = 29. Verify: 21² + 20² = 441 + 400 = 841 = 29². ✓
Question 2 Multiple Choice
A student uses the parametrization with m = 3, n = 1, computing a = 8, b = 6, c = 10, and verifies 8² + 6² = 10². But she notes that gcd(8, 6, 10) = 2, so the triple is not primitive. What went wrong?
AShe chose n = 1, which is not allowed in the parametrization for primitive triples
BShe used m and n of the same parity (both odd), which always produces an even-valued non-primitive triple
CShe computed b = 2mn incorrectly; the formula for primitive triples uses b = mn
DThe parametrization only works for m and n that are prime; m = 3 is prime but n = 1 is not
When m = 3 and n = 1 are both odd: a = m²−n² = 9−1 = 8 (even), b = 2mn = 6 (even), c = m²+n² = 9+1 = 10 (even). All three are even, so gcd ≥ 2 — not primitive. The parity condition (m and n of opposite parity) is precisely what prevents this: if m is even and n is odd (or vice versa), then m²−n² is odd, 2mn is divisible by 2 but not 4 in the primitive case, and m²+n² is odd, giving gcd(a, b, c) = 1. The condition gcd(m, n) = 1 alone is not enough to guarantee primitivity — opposite parity is independently required.
Question 3 True / False
The pair m = 3, n = 2 generates the primitive Pythagorean triple (5, 12, 13).
TTrue
FFalse
Answer: True
Verify: a = m²−n² = 9−4 = 5, b = 2mn = 2(3)(2) = 12, c = m²+n² = 9+4 = 13. Check: 5² + 12² = 25 + 144 = 169 = 13². ✓ Conditions: m = 3 (odd), n = 2 (even) — opposite parity; gcd(3, 2) = 1; m > n. All conditions satisfied, confirming a primitive triple.
Question 4 True / False
The parametrization a = m²−n², b = 2mn, c = m²+n² misses many primitive Pythagorean triples — for instance, triples where both legs a and b are odd.
TTrue
FFalse
Answer: False
The parametrization is COMPLETE: it produces every primitive Pythagorean triple exactly once (up to swapping a and b). In fact, in any primitive triple one leg is always even (b = 2mn is always even) and one is always odd — a primitive triple can never have two odd legs, because if a and b are both odd, then a² + b² ≡ 2 (mod 4), and no perfect square is ≡ 2 (mod 4). So there are no missing triples of this kind. Every primitive triple is accounted for by exactly one valid (m, n) pair.
Question 5 Short Answer
What does it mean for the parametrization of primitive Pythagorean triples to be 'complete,' and why is completeness a remarkable property for a Diophantine equation?
Think about your answer, then reveal below.
Model answer: Completeness means the parametrization produces every primitive Pythagorean triple and misses none — each triple appears exactly once for some valid (m, n) pair. This is remarkable because most Diophantine equations (integer polynomial equations) either have no solutions, finitely many solutions, or infinitely many solutions with no known closed-form description. A complete explicit parametrization — a finite formula that generates all solutions with no gaps and no duplicates — is rare. The Pythagorean equation is unusual in admitting such a parametrization, made possible by its multiplicative structure and the factorization in the Gaussian integers.
For comparison, the Fermat equation xⁿ + yⁿ = zⁿ has no nontrivial integer solutions for n > 2 (Fermat's Last Theorem), and the general cubic Diophantine equation has no complete parametrization. The Pythagorean case is special: the equation a² + b² = c² factors nicely over ℤ[i], allowing a complete description.