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Is there a known complete formalization of solutions to $a^2 + b^2 = c^2 + k$ for a fixed constant $k>0$ similar to the one for primitive Pythagorean triples (i.e. $(a,b,c) = (m^2-n^2,2mn,m^2+n^2)$ )?

I saw on-line somewhere that Euler found all solutions to $a^2+b^2=c^2+d^2$ (I think it was $(a,b,c,d) = (xy-wz,xz+wy,xy+wz,xz-wy)$) so if $k$ is a square number $k=r^2$ we can find $x,y,w,z$ such that $xz-wy = r$. However, I was wondering if there is some known formulation having only two parameters (like we have for Pythagorean triples) given that $k$ is fixed.

Ari
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  • http://math.stackexchange.com/questions/74931/integral-solutions-of-x2y21-z2 – individ Apr 18 '16 at 15:17
  • Looked into it, there is nothing with two parameters unless $k=0.$ Quickest is probably to take $k - b^2 = a^2 - c^2 = (a+c)(a-c),$ so iterate over all $b$ and whenever $k - b^2 \neq 2 \pmod 4$ find divisors $k-b^2 = uv$ with $u \equiv v \pmod 2.$ – Will Jagy May 03 '16 at 19:02
  • @WillJagy How were you able to prove that there is nothing with two parameters? – Ari May 04 '16 at 14:12
  • No proof, long, long experience. The most clear cut method for your problem is stereographic projection around one integer solution, that is easy enough, but is three parameters by nature. The fact that Pythagorean triples get down to just two parameters is a fortunate accident, explored fully by Fricke and Klein and related to Fuchsian groups. Meanwhile: your profile shows your principal interest in computers. Were you going to actually program this and work with the mathematics, or did you just want someone to assure you about low complexity? – Will Jagy May 04 '16 at 16:14
  • $(m^2-n^2)^2+(2mn)^2=m^4+n^4+2m^2n^2$, If $k=n^4+2m^2n^2$ then you have a parametric solution for $a^2+b^2=c^2+k$ – sirous Dec 16 '17 at 15:28
  • @WillJagy, can you give a link or reference for your comment on Fricke, Klein, and Fuchsian groups? I've been hacking with the relationship between Fuchsian groups and Pythagorean triples and would like some historical context. – brainjam Jan 08 '18 at 14:34
  • @brainjam the original book is Fricke and Klein (1897). The A.M.S. has just brought out an English translation. Much of the material you want is in Magnus, Noneuclidean Tesselations and their Groups. I found my own proof of the bit that I use over and over, http://math.stackexchange.com/questions/1972120/ternary-quadratic-forms/1976199#1976199 – Will Jagy Jan 08 '18 at 18:05
  • @brainjam here is the translation, seems publication has been moved back a couple of months http://bookstore.ams.org/ctm-3/ – Will Jagy Jan 08 '18 at 18:55
  • @WillJagy, I have Magnus, and nothing about Pythagorean triples jumps out at me. Which parts did you have in mind? Allen Hatcher's Topology of Numbers has some material in the first few chapters (Chapter 3 and earlier). – brainjam Jan 08 '18 at 19:02
  • @brainjam $$ \left( \begin{array}{rrr} 2 & 0 & 2 \ 2 & 2 & 2 \ 0 & 1 & 1\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & -1 \ \end{array} \right) \left( \begin{array}{rrr} 2 & 2 & 0\ 0 & 2 & 1\ 2 & 2 & 1 \ \end{array} \right) = 2 \cdot \left( \begin{array}{rrr} 0 & 0 & -1 \ 0 & 2 & 0 \ -1 & 0 & 0 \ \end{array} \right) $$ With that in mind, what are the vectors $(u,v,w)$ with $\gcd(u,v,w) = 1$ and $v^2 = w u ; ?$ – Will Jagy Jan 08 '18 at 19:19
  • here we go, on the right I should have used the familiar [1,0,-1; 0,2,0; 1,0,1] – Will Jagy Jan 08 '18 at 19:32
  • @brainjam $$ \left( \begin{array}{rrr} 1 & 0 & 1 \ 0 & 2 & 0 \ -1 & 0 & 1\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & -1 \ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & -1\ 0 & 2 & 0\ 1 & 0 & 1 \ \end{array} \right) = 2 \cdot \left( \begin{array}{rrr} 0 & 0 & -1 \ 0 & 2 & 0 \ -1 & 0 & 0 \ \end{array} \right) $$ – Will Jagy Jan 08 '18 at 19:42
  • the primitive vectors with $v^2 = w u$ are coprime parameters $s,t,$ then $u = s^2, $ $v = st,$ $w = t^2.$ If desired you can negate all three – Will Jagy Jan 08 '18 at 20:42

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I do not know of a solution to your first part unless $k-0$ if you want $k$ to be the same across the board. I also don't know about Euler finding all solutions to $a^2+b^2=c^2+d^2$ because there are an infinit number of them found most easily with deriviatives of Euclid's formula shown here as $$A=m^2-k^2,\quad B=2mk,\quad C=m^2+k^2\quad$$ By solving the C-function for k, we can find all triples with a given C by testing m-values to see which yield k-integers.

\begin{equation} 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 \end{equation} The lower limit ensures $m>k$ and the upper limit ensures $k\in\mathbb{N}$.

$$C=65\implies \bigg\lfloor\frac{ 1+\sqrt{130-1}}{2}\bigg\rfloor=6 \le m \le \lfloor\sqrt{65-1}\rfloor=8\quad\land \quad m\in\{7,8\}\Rightarrow k\in\{4,1\}\\$$ $$F(7,4)=(33,56,65)\qquad \qquad F(8,1)=(63,16,65) $$

Here we have $33^2+56^2=63^2+16^2$ and there are infinite numbers of these that can by found by testing C-candidates of the form where $C=4n+1$.

poetasis
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