Question

Let `D= a^2+b^2+c^2` where a and b are successive positive integers and `c=ab`.How will you show that `sqrtD` is an odd positive integer?

Answer 1

`D = (a^2+a+1)^2` which is the square of an odd integer.

Explanation:

Given `a`, we have:

`b = a + 1`

`c = ab = a(a+1)`

So:

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

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

`=a^4+2a^3+3a^2+2a+1`

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

If `a` is odd then so is `a^2` and hence `a^2+a+1` is odd.

If `a` is even then so is `a^2` and hence `a^2+a+1` is odd.