# 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?

Jun 28, 2016

See below

#### Explanation:

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

${a}^{2} + {b}^{2} + {a}^{2} {b}^{2} = {n}^{2} + {\left(n + 1\right)}^{2} + {n}^{2} {\left(n + 1\right)}^{2}$

which gives

$1 + 2 n + 3 {n}^{2} + 2 {n}^{3} + {n}^{4}$

but

$1 + 2 n + 3 {n}^{2} + 2 {n}^{3} + {n}^{4} = {\left(1 + n + {n}^{2}\right)}^{2}$

which is the square of an odd integer