Question

If a and b are natural number and a>b then show that(a*a+b*b),(a*a-b*b),(2ab) is a Pythagorean triplet .Find two Pythagorean triplets using any convenient values of a and b
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Answer 1

According to Pythagoras theorem in a right angled triangle,

p² + b² = h²

where p = perpendicular

b = base

h = hypotenuse

The given triplets are (a² + b²), (a²-b²) and 2ab

squaring these triplets we get,

1st = (a² + b²)² = a⁴ + b⁴ + 2a²b²

2nd = (a² - b²)² = a⁴ + b⁴ - 2a²b²

3rd = (2ab)² = 4a²b²

Adding the second and third triplet's square we get

= a⁴ + b⁴ - 2a²b² + 4a²b²

= a⁴ + b⁴ + 2a²b²

= (a² + b²)²

which is the square of the 1st triplet

Hence these triplets are Pythagorean triplets(proved)

Since these triplets are Pythagorean triplets , so multiplying any constant to all of them will also be a Pythagorean triplets

Hence the other pairs of Pythagorean triplets are:

2(a² + b²), 2(a²-b²) , 4ab

and

3(a² + b²), 3(a²-b²), 6ab

Answer 2

Answer:

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