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 · 2016-06-28T21:45:09

See below

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

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?