Athena F.

asked • 05/15/21

Using the polynomial identity (x^2+y^2)=(x^2-y^2)^2+(2xy)^2 , determine the Pythagorean Triple when x=3 and y=5 .

A) 3,4,5

B) 256.900.1156

C) 5,12,13

D) 16,30,34

Inactive Tutor

The identity that you present is FALSE. Review your post for accuracy.
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05/15/21

Tom K.

tutor
Mark is right, the left hand side should be squared. We get the Pythagorean triple 16-30-34; the one thing that must be understood is that when we calculate, we get -16, not 16. However, as the term is squared, it gives the same value as 16.
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05/16/21

Inactive Tutor

Pythagorean triples are defined only for positive integers. Therefore -16 is not allowed.
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05/16/21

1 Expert Answer

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Inactive Tutor answered • 05/17/21

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Hi Athena F.,


The Pythagorean triple Identity is: (x2 + y2)2 = (x2 - y2)2 + (2xy)2, where c = x2 + y2, a = x2 - y2, and b = 2xy. With x = 3 and y = 5:

a = 32 - 52 = 16, (this is -16, by the rules x should be greater than y (x > y), in this case √[(x2 - y2)2] = |a|).

b = 2*3*5 = 30

c = 32 + 52 = 34


I hope this helps, Joe.

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