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.
So D is a perfect square and an integer.