Question

find the pathagorean triplets ,one of whose natural number is three

Answer 1

The triplet is 3, 4 and 5

Answer 2

How do I find pythagorean triplets with one given number?

If one of the sides has odd value like a = 11, then the other side ‘b' has to be even, hypotenuse = h

& a² + b² = h²

=> 11² = h² - b²

=> 121 = (h+b) (h-b)

=> 121 = 121 x 1 ( if a is odd, we take h- b = 1, & if a is even h-b is taken 2)

Now, since h- b = 1 , so h & b are consecutive numbers.

But, h+ b = 121 , so 2 consecutive numbers the sum of which is 121, can be found out by dividing (121 -1) by 2 . So, the consecutive numbers are 60 & 61

Hence pythagorean triplets will be 11, 60 & 61 ………………(1)

& 61² = 11² + 60²

=> 3721 = 121 + 3600 ( verified)

Now, if a = even number 14, then b = even , hypotenuse = h

h² = 14² + b²

=> 196 = h² - b² = ( h+b) ( h- b) = 98 x 2

Here, b is obtained by dividing ( 98 -2) by 2

=> 196 = ( 50 + 48) ( 50 - 48)

Hence, triplet has to be 14, 48 & 50 ………….(2)

50² = 14² + 48²

=> 2500 = 196 + 2304 = 2500 ( verified)

Hope this will help you.. ✌