1. ## Sums of squares

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.

2. 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.

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

3. 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.

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

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:

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

(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.

6. 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

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

7. ## Responding to Bruno J.

"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.

8. 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... )

9. ## Thank you Defunkt

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.

10. ## As promised

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

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).

11. 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).

12. ## Drexel

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?)

13. 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.