Let $D = a^{2} + b^{2} + c^{2}$ $D= a^2+b^2+c^2$ where a and b are successive positive integers and $c = a b$ $c=ab$ . How will you show that $\sqrt{D}$ $sqrtD$ is an odd positive integer?

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Answer 1 · 2016-06-28T21:45:09

See below

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

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

which gives

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

but

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

which is the square of an odd integer

Let D=a2+b2+c2D= a^2+b^2+c^2 where a and b are successive positive integers and c=abc=ab. How will you show that √DsqrtD is an odd positive integer?