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I have found that the integer solution of $a^2+b^2=c^2+d^2$ is $(a,b,c,d)=(pr+qs,ps-qr,pr-qs,ps+qr)$ for integer $p,r,q,s$.

I wonder if there is an explicit relation between $a,b,c,$ and $d$? Or could you give me a hint on what topics should I learn to find this relation if it exists?

Thank you for all your help.

ZeroToZero
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    FYI, the $(a,b,c,d)=(pr+qs,ps-qr,pr-qs,ps+qr)$ solution you found is basically the Brahmagupta–Fibonacci identity. – John Omielan Sep 09 '22 at 10:03
  • you can start with any odd number (that is not a prime) to produce $a^2+b^2=c^2+d^2$. For example $15=4^2-1^2=35$ but also $15=8^2-7^2=151$. Since $15=15$, then $4^2-1^2=8^2-7^2$ or $4^2+7^2=1^2+8^2=65$. The relation between $a,b$ is that of factors $a+b$ and $a-b$ of $N=a^2+b^2$ and the same with $c,d$. But it's not clear to me how $15$ and $65$ are related (except by the fact that they share a factor in this case. – user25406 Sep 09 '22 at 11:01
  • @ZeroToZero Have any of these contributions been helpful? – poetasis Sep 10 '22 at 00:42
  • Maybe this site will help you: https://sites.google.com/site/tpiezas/003 (#10. Form $mx^2+ny^2=mz^2+nt^2$) – RDK Sep 10 '22 at 02:05
  • https://math.stackexchange.com/questions/153603/diophantine-equation-a2b2-c2d2/736164#736164 – individ Sep 10 '22 at 07:26
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    Thank you @JohnOmielan. I don't know the name of the identity previously. – ZeroToZero Sep 10 '22 at 14:56
  • Thank you, @user25406 for your idea – ZeroToZero Sep 10 '22 at 14:57
  • Thank you @RDK. Unfortunately, I think it's still the Brahmagupta solution that doesn't relate explicitly the a, b, c, and d. – ZeroToZero Sep 10 '22 at 15:00
  • Thank you, @individ. But it still doesn't relate a, b, c, and d explicitly. – ZeroToZero Sep 10 '22 at 15:01

2 Answers2

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I do not know of any relationship between the summands, only the relationship between the sum and it's factors.

For Pythagorean triples, $\space (A^2+B^2=C^2),\space$ there are $\space 2^{n-1}\space$ primitive triples where $\space n\space$ is the number of distinct prime factors of $\space C.\quad$ A primitive triple is one where $\space GCD(A,B,C)=1.\quad $ For example, the number $\space65=5*13\space$ so there are $\space 2^{2-1}=2^1=2\space$ primitive triples with that hypotenuse value. To find them, we begin with Euclid's formula shown here as:

$$A=m^2-k^2 \quad B=2mk \quad C=m^2+k^2$$ and, solving for $\space k,\space$ we test a defined range of $\space m$-values to see which, if any, yield integers.

$$C=m^2+k^2\implies k=\sqrt{C-m^2}\\ \qquad\text{for}\qquad \bigg\lfloor\frac{ 1+\sqrt{2C-1}}{2}\bigg\rfloor \le m \le \lfloor\sqrt{C-1}\rfloor$$ The lower limit ensures $m>k$ and the upper limit ensures $k\in\mathbb{N}$. $$C=65\implies \bigg\lfloor\frac{ 1+\sqrt{130-1}}{2}\bigg\rfloor=6 \le m \le \lfloor\sqrt{65-1}\rfloor=8\quad \\ \text{and we find} \quad m\in\{7,8\}\longrightarrow k\in\{4,1\}\\$$ $$F(7,4)=(33,56,65)\qquad F(8,1)=(63,16,65) $$

So $$33^2+56^2=63^2+16^2=65^2$$

$\textbf{Update}"\quad$ The parts of a Pythagorean triple have characteristics that may or may not be called relationships.

  1. Side-$A\space$ can be any odd number greater than one.
  2. Side-$B\space$ can be any multiple of four.
  3. Side-$C\space$ must be of the form $\space 4x+1, x\in\mathbb{N}\space$ but not all such values are valid. A list of the first couple of dozen hypotenuse values is shown here.
poetasis
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From https://sites.google.com/site/tpiezas/003 (10. Form $mx^2+ny^2=mz^2+nt^2$-S. Realis(complete):

$$ u^2+nv^2=x^2+ny^2 \\ \Leftrightarrow \small \{ u, v, x, y \}=\{ a^2-n(a-b)^2+n(a-c)^2, b^2-(a-b)^2+n(b-c)^2, a^2+nb^2-nc^2, c^2-(a-c)^2-n(b-c)^2 \}. $$

Substituting $n=1$;

$$ u^2+v^2=x^2+y^2 \\ \Leftrightarrow \small \{ u, v, x, y \}=\{ a^2-(a-b)^2+(a-c)^2, b^2-(a-b)^2+(b-c)^2, a^2+b^2-c^2, c^2-(a-c)^2-(b-c)^2 \}. $$

So, this is the generalized form;

$\small \{a, b, c, d\}=\{m^2-(m-n)^2+(m-l)^2, n^2-(m-n)^2+(n-l)^2, m^2+n^2-l^2, l^2-(m-l)^2-(n-l)^2\}.$

RDK
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