Question

If a^2+b^2+c^2=d, where a and b are consecutive positive integers and c=ab then sqrt(D) is?

• Sometimes rational
• Sometimes odd
• Always odd

It is given a, b are consecutive numbers so without the loss generality we can assume b=a+1

So the given equation become D=a^2+(a+1)^2+a^2(1+a)^2

a^2(1+a)^2 is always even. And exactly one of a^2, (1+a)^2 is even. So D is odd. Which implies sqrtD will be an odd if it is an integer.

Now let us try to figure if it is perfect square.

D=a^2+(a+1)^2+a^2(1+a)^2

=2a(a+1)+1+a^2(1+a)^2

=(a(1+a)+1)^2

So D is a perfect square and an integer.