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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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 · 2016-06-20T11:47:16

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

Explanation:

Given $a$ $a$ , we have:

$b = a + 1$ $b = a + 1$

$c = a b = a (a + 1)$ $c = ab = a(a+1)$

So:

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

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

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

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

If $a$ $a$ is odd then so is $a^{2}$ $a^2$ and hence $a^{2} + a + 1$ $a^2+a+1$ is odd.

If $a$ $a$ is even then so is $a^{2}$ $a^2$ and hence $a^{2} + a + 1$ $a^2+a+1$ is odd.

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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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Explanation:

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

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Explanation:

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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?