# 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 20, 2016

$D = {\left({a}^{2} + a + 1\right)}^{2}$ which is the square of an odd integer.

#### Explanation:

Given $a$, we have:

$b = a + 1$

$c = a b = a \left(a + 1\right)$

So:

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

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

$= {a}^{4} + 2 {a}^{3} + 3 {a}^{2} + 2 a + 1$

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

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.