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If it is only one quadratic equation $x^2-xy+y^2=a^2$, we can get some integral solutions as follows. \begin{align*} &\left\{ \begin{split} x&=k(2mn-n^2)\\ y&=k(m^2-n^2)\\ a&=k(m^2-mn+n^2)\\ \end{split}\right. &\quad \left\{ \begin{split} x&=k(m^2-2mn)\\ y&=k(2mn-n^2)\\ a&=k(m^2-mn+n^2)\\ \end{split}\right.\\ \\ &\qquad\qquad\downarrow p=m-n&p=m-n\downarrow\qquad\qquad\qquad\\ \\ &\left\{ \begin{split} x&=k(2np+n^2)\\ y&=k(2np+p^2)\\ a&=k(p^2+np+n^2)\\ \end{split}\right. &\quad \left\{ \begin{split} x&=k(p^2-n^2)\\ y&=k(2np+n^2)\\ a&=k(p^2+np+n^2)\\ \end{split}\right.\\ \end{align*}

However, I'm not sure if these are the complete solutions of Diophantine equation $x^2-xy+y^2=a^2$.

But how to solve this Diophantine equation system in integers? \begin{align*} \left\{ \begin{split} \large{x^2-xy+y^2}&\large{=a^2}\\ \large{y^2-yz+z^2}&\large{=b^2}\\ \large{x^2-xz+z^2}&\large{=c^2}\\ \end{split}\right. \end{align*} I got some non-trivial examples:

$ \begin{align*} \left\{ \begin{split} x&=\phantom{0}7\\ y&=15\\ z&=40\\ a&=13\\ b&=35\\ c&=37 \end{split}\right. \end{align*}$,$\begin{align*} \left\{ \begin{split} x&=\phantom{0}21\\ y&=\phantom{0}56\\ z&=120\\ a&=\phantom{0}49\\ b&=104\\ c&=111 \end{split}\right. \end{align*}$,$\begin{align*} \left\{ \begin{split} x&=\phantom{0}77\\ y&=117\\ z&=165\\ a&=103\\ b&=147\\ c&=143 \end{split}\right. \end{align*}$.

https://benvitalenum3ers.wordpress.com/2016/12/20/make-x2-xy-y2-x2-xz-z2-y2-yz-z2-squares-part-4/

Eufisky
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  • HINT Consider symmetry $ (x,y,z) \rightarrow \sqrt{\dfrac{xy+yz+zx}{3}}$ – Narasimham Mar 13 '20 at 14:46
  • Another hint: costruct point $X$ and draw points $A$, $B$, $C$ such that $\angle AXB = \angle BXC = \angle CXA = 120 °$. Then sides of triangle $ABC$ are $a$, $b$ and $c$ with $X$ be a Fermat point ( https://en.wikipedia.org/wiki/Fermat_point). Now use the property of Heronian triangle (https://en.wikipedia.org/wiki/Heronian_triangle). – Daniyar Aubekerov Mar 13 '20 at 15:10
  • emmmm~. According to your method, three equal angles should also be $60^\circ$. @DaniyarAubekerov $\dfrac{x^2+y^2-a^2}{2xy}=\dfrac{1}{2}=\cos60^\circ$ – Eufisky Mar 13 '20 at 16:16
  • What is $x^2-xy+y^2=a^2$? an equation with 2 variables x,y or an equation with 3 variables x,y,a? – miracle173 Mar 16 '20 at 06:09
  • Thank you for your remind! @miracle173 there was a clerical error. – Eufisky Mar 16 '20 at 14:48

1 Answers1

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Dickson book (history of theory of numbers) volume 2 page 511 has solution.

$x=(n^2-1)(m^2-1)$

$y=(2n-1)(m^2-1)$

$z=(n^2-1)(2m-1)$

Where, $m=2×(2q^2-pq-qv)/(3q^2-2pv+pq-2p^2)$

$(p,q,v)=((2n-1),(n^2-1),(n^2-n+1))$

For n=3, $(p,q,r)=(5,8,7)$

& $m=(4/7)$

$(x,y,z)=((-264),(-165),(56))$

$(a,b,c)=(231,199,296)$

Robert
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