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

See below

Explanation:

Making `a=n` and `b = n+1` and substituting in

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

which gives

`1 + 2 n + 3 n^2 + 2 n^3 + n^4`

but

`1 + 2 n + 3 n^2 + 2 n^3 + n^4 = (1 + n + n^2)^2`

which is the square of an odd integer