How to prove Sum of squares of $n$ numbers is unique

Asked Jul 20 '13 at 13:38

Active Jul 20 '13 at 13:40

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I was solving a programming problem where I needed to prove that sum of squares of $n$ numbers is unique. i. e. if calculate the sum of squares of equal number of positive integers then if they are same it means that all numbers are the same.

How do I prove it?

The range of numbers is between 0 and 127(inclusive)
or does the range matter?

edited Jul 20 '13 at 13:40

Hagen von Eitzen

374,180

asked Jul 20 '13 at 13:38

tapananand

$$S= \frac{n(n+1)(2n+1)}6$$ is an increasing function for $n>0$ – lab bhattacharjee Jul 20 '13 at 13:40
5

But $7^2+1^2=5^2+5^2$ ! – Hagen von Eitzen Jul 20 '13 at 13:41
@HagenvonEitzen but can you find an example with distinct numbers on both sides? – Harald Hanche-Olsen Jul 20 '13 at 13:42
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Or even $3^2+4^2 = 5^2+0^2$ – TZakrevskiy Jul 20 '13 at 13:42
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Sure, $11^2+2^2=10^2+5^2$ or $7^2+4^2=8^2+1^2$ – Hagen von Eitzen Jul 20 '13 at 13:43
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@HagenvonEitzen ah, that is more satisfying. And now I see how such examples are easily found: $a^2+b^2=c^2+d^2$ is equivalent to $(a-c)(a+c)=(d-b)(b+d)$, so you just look for two different factorizations of a number. – Harald Hanche-Olsen Jul 20 '13 at 13:44
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Find here(http://mathworld.wolfram.com/SumofSquaresFunction.html) for the number of representations of an sum of squares – lab bhattacharjee Jul 20 '13 at 13:49
By the way, I found these examples by looking at complex number multiplications $(a+bi)(c\pm di)$, e.g. $(1+2i)(2\pm3i)$ for the last example. – Hagen von Eitzen Jul 20 '13 at 13:50
If $a^2+b^2=c^2+d^2$ then $a^2-c^2=d^2-b^2$ so that there is a number $n=(a+c)(a-c)=(d+b)(d-b)$. So take a number $n$ and factorise as $n=pq=rs$. We need $p-q$ even and $r-s$ even. Then $a=\frac{p+q}2$, $c=\frac{p-q}2$, $d=\frac{r+s}2$, $b=\frac{r-s}2$ will do. – Mark Bennet Jul 20 '13 at 13:50
So this means the solution that i saw was wrong. I'm sorry – tapananand Jul 20 '13 at 13:51

How to prove Sum of squares of $n$ numbers is unique

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