Prove that there are infinitely many positive integers n such n(n+1) can b expressed as a sum of two positive squares in at least two different ways.(a^2 + b^2 and b^2 + a^2 are considered as the same representation.....

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- Sep 16th 2007, 09:50 AMblah_blaholymiad question....
Prove that there are infinitely many positive integers n such n(n+1) can b expressed as a sum of two positive squares in at least two different ways.(a^2 + b^2 and b^2 + a^2 are considered as the same representation.....

- Sep 16th 2007, 10:16 AMThePerfectHacker
There are infinitely many squares $\displaystyle x$ so that $\displaystyle 3x-1=y^2$ because if $\displaystyle x=z^2$ we have $\displaystyle 3z^2 - 1=y^2$ which is Pellian with infinitely many solutions.

Let $\displaystyle n$ be a number with the condition that $\displaystyle n$ is a square**and**$\displaystyle 3n-1$ is a square. Which we know there are infinitely many.

Then,

$\displaystyle n(n+1) = a^2 + b^2$

Where $\displaystyle a=n \mbox{ and }b=\sqrt{n}$.

And another,

$\displaystyle a=(n-1) \mbox{ and }b=\sqrt{3n-1}$. - Sep 18th 2007, 07:41 AMblah_blah
dint really understand.....can u simplify n explain it??

- Sep 20th 2007, 07:38 PMThePerfectHacker
If you find such an $\displaystyle n$ so that it is a square

**and**$\displaystyle 3n-1$ is a square. Then $\displaystyle \sqrt{n}$ and $\displaystyle \sqrt{3n-1}$ are integers.

This means,

$\displaystyle n(n+1) = n^2 + n = (n)^2 + (\sqrt{n})^2$

And,

$\displaystyle n(n+1) = n^2 + n = (n^2 -2n + 1) + (3n-1) = (n-1)^2 + (\sqrt{3n-1})^2$.

So the question is. "Are there infinite many such $\displaystyle n$ having this property?". The answer is yes. This is a consequence of the Pellian equation is you really want to show it. - Sep 22nd 2007, 07:18 AMblah_blah
thanks...i understood d whole thing except 4 d pell equation...i tried goin thru wolfram math world's explanation 4 it but i din understand it....can u help??