Sums of squares

**Defunkt** · January 3rd 2010, 11:49 AM

Hi all. A friend asked me an interesting question about representations of integers as sums of two squares: what could be a condition, given $n \in \mathbb{N}$ , for n to have two different representations (ie. two different couples $(a,b) \neq (c,d) \in \mathbb{N \times N}$ such that $n=a^2+b^2=c^2+d^2$ ) as a sum of two squares? I was wondering if someone could help!

I haven't much knowledge (yet) in number theory or fields/rings (but I did study some about groups), but feel free to use any definitions and I'll just look them up. Any direction would also be nice (instead of a solution).

Happy new year, and again thanks for any help.

**Soroban** · January 3rd 2010, 12:49 PM

Hello, Defunkt!

A friend asked me an interesting question about representations of integers as sums of two squares:
what could be a condition, given $n \in \mathbb{N}$ ,
for $n$ to have two different representations as a sum of two squares?

Lucky for us, the problem has been solved long ago.

$n$ must be the product of two sums-of-squares.

. . $n \;=\;(a^2+b^2)(c^2+d^2) \;=\;\begin{Bmatrix}(ac+bd)^2 + (ad-bc)^2 \\ \\[-3mm] (ac-bd)^2 + (ad + bc)^2 \end{Bmatrix}$

**Defunkt** · January 3rd 2010, 01:10 PM

Originally Posted by Soroban

Hello, Defunkt!

Lucky for us, the problem has been solved long ago.

$n$ must be the product of two sums-of-squares.

. . $n \;=\;(a^2+b^2)(c^2+d^2) \;=\;\begin{Bmatrix}(ac+bd)^2 + (ad-bc)^2 \\ \\[-3mm] (ac-bd)^2 + (ad + bc)^2 \end{Bmatrix}$

Hi Soroban! Thank you for the quick answer.

Do you know of any source (website, book) that deals with these sort of problems specifically? I would like to read some about it myself... this seems like an interesting topic.

**Bruno J.** · January 3rd 2010, 03:38 PM

I suggest Hardy and Wright's The Theory of Numbers. The topic is covered extensively.

**wonderboy1953** · January 4th 2010, 11:10 AM

Hello Defunkt.

To answer your question about a natural number having two or more non-trivial sum of squares representations, I believe I read that such representations don't exist.

However you may be interested in the following:

Multigrade example (through to the fifth power, a pentagrade):

$1^n +15^n + 22^n + 50^n + 57^n + 71^n$ = $2^n + 11^n + 27^n + 45^n + 61^n + 70^n$ = $5^n + 6^n + 35^n + 37^n + 66^n + 67^n$ ; $n = 1,2,3,4,5$

(my understanding is that this type of multigrade can have infinite equal signs).

If you want to pursue this further, you can try a Google search under number theory - multigrades.

**Bruno J.** · January 4th 2010, 03:41 PM

Originally Posted by wonderboy1953

Hello Defunkt.

To answer your question about a natural number having two or more non-trivial sum of squares representations, I believe I read that such representations don't exist.

Dude, what are you talking about? Why do you keep bringing up "multigrades" when they are not in any obvious way related to Defunkt's question, all while feeding him false statements. Read Soroban's post.

Anyways. Let $U(n)$ be the number of representations of $n$ as a sum of two squares. (We count $a^2+b^2$ and $b^2+a^2$ as different representations if $a \neq b$ .) Then we have the beautiful theorem

$U(n) = 4\sum_{d|n,\ d\equiv 1 \mod 2}(-1)^\frac{d-1}{2}.$

**wonderboy1953** · January 5th 2010, 09:08 AM

"Dude, what are you talking about? Why do you keep bringing up "multigrades" when they are not in any obvious way related to Defunkt's question, all while feeding him false statements. Read Soroban's post."

I did. Defunkt is talking about representing as sum of two squares with two numbers ( $a^2 + b^2$ or $c^2 + d^2$ while Soroban is talking about a product leading to the sum of two squares, each square however being composed of four numbers which isn't the same as what Defunkt brought up.

The multigrade example I brought up does have some relationship to what Defunkt brought up and I feel would be of interest to him (I never said that the example answers his question and I did say that I don't believe that there are two, non-trivial, different two-termed sum of squares representation for any positive integer).

Since this is Defunkt's thread then he should have the final say on this matter.

**Defunkt** · January 5th 2010, 09:28 AM

Well, Soroban did provide a sufficient condition for a number to have two representations as a sum of two squares. Wonderboy, if you want a counter example to your claim:

$5\cdot 10^4 = 40^2 + 220^2 = 100^2 + 200^2 = 152^2 + 164^2$

But thanks regardless for trying to help. I am more interested in sums of squares rather than multigrades (maybe eventually I'll have time to look at that as well...

)

**wonderboy1953** · January 5th 2010, 09:46 AM

Not exactly a claim, but a belief.

I have at home a three-termed sum of squares with two equal signs which I'll post in tomorrow for all to study.

**wonderboy1953** · January 5th 2010, 03:59 PM

$9^n + 25^n + 26^n = 10^n + 21^n + 29^n = 11^n + 19^n + 30^n$

This is a bigrade. You can let n = 2 to get sum of squares (as a note I do distinctly remember it being reported a multigrade can't have less than three nonzero terms which is why I believed that the proposed problem didn't have a solution. I stand corrected).

**Drexel28** · January 6th 2010, 06:07 PM

Originally Posted by wonderboy1953

$9^n + 25^n + 26^n = 10^n + 21^n + 29^n = 11^n + 19^n + 30^n$

This is a bigrade. You can let n = 2 to get sum of squares (as a note I do distinctly remember it being reported a multigrade can't have less than three nonzero terms which is why I believed that the proposed problem didn't have a solution. I stand corrected).

What is with your absolute obsession with "multigrades"?

**wonderboy1953** · January 7th 2010, 02:48 PM

I have posted other subjects (e.g. see my postings at the math challenge problems section). For me multigrades hold a fascination and visitors have the option of ignoring what I've posted [on this would it help you for me to list multigrade(s) in the title so you can skip my post?)

**Drexel28** · January 7th 2010, 03:12 PM

Originally Posted by wonderboy1953

I have posted other subjects (e.g. see my postings at the math challenge problems section). For me multigrades hold a fascination and visitors have the option of ignoring what I've posted [on this would it help you for me to list multigrade(s) in the title so you can skip my post?)

No, I have no problem with your fascination with multi-grades. I was just wondering what specific, if any, appeal they held to you.

**wonderboy1953** · January 7th 2010, 03:17 PM

Check out my latest post under the Riemann Manifold thread that I did today
in the number theory section.

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Responding to Bruno J.

Thank you Defunkt

As promised

Drexel

This might help you Drexel

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