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?

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Answer 1 · 2016-06-20T11:47:16

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

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.

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