Find Pythagorean triplets with one among them 16
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Answer: For any natural number m > 1, we have (2m) ² + (m2 – 1)²= (m2 + 1)² . So, 2m, m² – 1 and m² + 1 forms a Pythagorean triplet.
Let 2 m = 16,
m = 8
m² + 1 = 8² + 1 = 64 + 1 = 65
m²- 1 = 8 2 - 1 = 64 - 1 = 63
check:
16² + 63² = 256 + 3969 = 4225 = 65²
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