Question

A and B are natural numbers and A is greater than B then show that a square + b square, A square minus b square , 2ab is a Pythagorean triplet find two pythagorean triplets using any convenient values of a and b​

Answer 1

The given numbers a²+b², a²-b², 2ab form an Pythagorean triplet

A Pythagorean Triplet of 3 numbers a,b,c is one which satisfies the identity,

a² + b² = c²

Given numbers are a²+b², a²-b², 2ab

Taking the first term and squaring it.

(a²+b²)² = (a²)² + (b²)² + 2*a²*b² = a⁴ + b⁴ + 2a²b²

(a²+b²)² = a⁴ + b⁴ - 2a²b² + 4a²b²

(a²+b²)² = (a² - b²)² + 4a²b²

(a²+b²)² = (a² - b²)² + (2ab)²

Hence, we can say that the given terms make a Pythagorean Triplet

Answer 2

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