Question

If the sum of the squares of three numbers is equal to the square of their sum, prove that the
sum of the products of the three numbers taking two at a time is equal to zero.
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Answer 1

Well, let us suppose three natural numbers i.e. a,b and c. And choose c such that c is a multiple of a and b. Then the relation will be

a^2+b^2+(a.b)^2=x^2. Here x can be written as a*b+1. so the given equation becomes.

a^2+b^2+(a*b)^2=[(a*b)+1]^2. And a-b= +- 1.

For e.g.

2^2+3^2+6^2=7^2

3^2+4^2+12^2=13^2

and so on...