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I am having problem in solving this question:

If a number $a$ is equal to sum of square of two different natural numbers then $2a$ is also equal to sum of square of two unequal natural numbers.

I cannot find my way through it.

Please help.

  • 3
    Hint: In general, if $n,m$ are both sums of two squares then so is $mn$ since $(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2$. – 2017-01-01

4 Answers 4

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Yes, It is true.

Hint: $$a=n^2+(n+m)^2$$ $$2a=2n^2+2(n+m)^2=2n^2+2(n^2+m^2+2mn)=(2n+m)^2+m^2$$

  • 0
    How did you come by this answer? – 2017-01-01
  • 5
    @N.S.JHON I just took a as it is looking and arranged the 2a to form 2 perfect squares – 2017-01-01
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A solution using Gaussian integers, which perhaps clarifies a point.

By assumption $a = x^2 + y^2 = (x + i y) (x - i y)$.

Then $$ 2 a = (1 + i) (1 -i) (x + i y) (x - i y) = ((x-y) + i (x+y)) ((x-y) - i (x+y)), $$ so that $$ 2 a = (x-y)^2 + (x+y)^2 $$ and now use the fact that $x \ne y$.

  • 2
    Definitely, and I know I was risking a downvote, but it's just one more answer, which can be safely ignored. – 2017-01-01
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If $x = y^2 + z^2 \implies 2x = (y+z)^2 + (y-z)^2$.

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Let $b>c$. Given

$\ \ \ \ \ a=b^2+c^2$

$\Rightarrow2a=2b^2+2c^2$

$\Rightarrow2a=b^2+c^2+2bc+b^2+c^2-2bc$

$\Rightarrow2a=(b+c)^2+(b-c)^2$

Hence, it is true.