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I'm working in a difficult functional equation, and I have reduced the problem to the following question ($\mathbb{N}$ denotes the set of non negative integers $0,1,2,3,4\cdots$)

Question:

Can we find all sequences of non negative integers ($\forall i\in \mathbb{N} \,\, a_i\in \mathbb{N}$) such that $$\begin{align} \forall i,j,k,l\in\mathbb{N} && i^2+j^2=k^2+l^2 &\Rightarrow a_i^2+a_j^2=a_k^2+a_l^2\end{align}$$

My try: I know the Euler's complete solution to the Diophantine equation :

$$x^2+y^2=z^2+t^2$$ $(ab+cd,ac-bd,ab-cd,ac+bd)$ but this is not very useful.

I also tried another way which is if we suppose that the sequence is injective ($a_i\neq a_j$ if $i\neq j$, maybe later we can prove that either the sequence is constant or this will be true), and under this assumption we can deduce that if $a_i+a_j$ is odd then: $$ d(|a_i^2-a_j^2|)\geq |i^2-j^2| \\ r_2(a_i^2+a_j^2)\geq r_2(i^2+j^2)$$

but the problem is we don't know much about the function $d(n)$ the number of divisors of $n$ and $r_2(n)$ the number of ways to write $n$ as the sum of two squares.

Expected solution: The only sequences I found are $a_i=c$ or $a_i=ki$ But is it the only sequences that fits our needs?, can we find a sequence with two values $c$ and $c'$ that fits the assertion?

Any comments, suggestions, any consequences of such assertion are welcome,Thank you for you help.

Bart Michels
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Elaqqad
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    Tiny note: you don't need your multiplier in the linear solution to be square - $a_i=ki$ works for any $k$. Also, it would help to clarify whether you're allowing zero values for $i/j/k/l$; you mention 'non-negative' for the $a_*$ but that doesn't say anything about the indices. Other than that, interesting question! – Steven Stadnicki Feb 24 '15 at 00:33
  • The question is not entirely clear. Can write more formally? What is the relationship between the numbers in the system ? – individ Feb 24 '15 at 04:16
  • @StevenStadnicki, I add the some improvement to the question, and I corrected the the linear solution, for the $indexes$ 0 is allowed, e.g : $a_0^2+a_5^2=a_3^2+a_4^2$, the problem when I write equations like this is: a sigle $a_i$ may verify infinity many equations, but for two elements $a_i$ and $a_j$ the number of such equation that contains $a_i$ and $a_j$ is finite (which allowed_me to conclude the two equations above). – Elaqqad Feb 24 '15 at 11:37
  • Can it is necessary to solve the following system of equations? $$\left{\begin{aligned}& i^2+j^2=k^2+l^2\&(a+ib)^2+(a+jb)^2=(a+kb)^2+(a+lb)^2\end{aligned}\right.$$ – individ Feb 24 '15 at 17:16
  • This system of equations is equivalent to $i+j=k+l$ so ${i,j}={k,l}$ – Elaqqad Feb 24 '15 at 20:11
  • @Elaqqad What was the original functional equation? – user574848 May 24 '19 at 10:00
  • @user574848 : I really don't remember the original equation. I don't think I have solved it . – Elaqqad Jun 26 '19 at 18:17
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    This question has already been answered here – Ewan Delanoy Jul 29 '23 at 10:29
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    Let $r$, $s$ and $t$ in $\mathbb{N}$. Then the sequence $(a_n)$ defined by $$a_n=\left{\begin{array}{ccc}r&\text{if}&n\equiv0;[4]\s&\text{if}&n\equiv1;[2]\t&\text{if}&n\equiv2;[4]\end{array}\right.$$ is a solution which is not of the type expected by the author of the initial message.

    To prove that $$\begin{align} \forall i,j,k,l\in\mathbb{N} && i^2+j^2=k^2+l^2 &\Rightarrow a_i^2+a_j^2=a_k^2+a_l^2\end{align},,$$ simply reason mod 8.

     – uvdose Aug 07 '23 at 18:34
  • @EwanDelanoy Yes indeed I was not aware of your question. If you write a simple answer I will accept it. – Elaqqad Dec 26 '23 at 00:32

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