How to solve the Diophantine equation：$x^2+xy+y^2=r^2,x,y,r\in \Bbb Z$

Question

I know this equation has trivial solutions: (x,y)=(±r,0),(0,±r),(r,-r),(-r,r) actually, what I want to ask is： what does r equals,this equation have nontrivial solutions? $$x^2+xy+y^2=r^2,x,y,r\in \Bbb Z$$

Answer 1

All primitive $\gcd(x,y,r) = 1$ solutions come from $$ x = u^2 - v^2, $$ $$ y = 2uv + v^2, $$ $$ r = u^2 + uv + v^2 $$

Answer 2

Your question belongs to a category which could be called "normic equations", and whose solutions can be entirely parametrized when the "norm form" is amenable enough. "Norm form" here means a homogeneous polynomial expressing (with adequate coordinates) the norm map $N$ of a number field $K/\mathbf Q$. Here $a^2+ab + b^2= (a-bj)(a-bj^2)=N(a-bj)$ where $K=\mathbf Q(j)$, with $ 1+j+j^2=0$, and your original diophantine equation $x^2+xy+y^2=r^2$ can obviously be written as $N(a-bj)=1$. Hilbert's thm. 90 for the cyclic extension $K/\mathbf Q$ (or direct computation) asserts that ($a-bj$) is the quotient of two conjugate elements of $K$, say $a-bj= (u-vj)/(u-vj^2)=(u-vj)^2/N(u-vj)$. By developping and identifying you get the parametrization of the primitive solutions $(x,y,r)$ given by @Will Jagy, namely $(u^2-v^2, 2uv+v^2, u^2+uv+v^2)$.

Geometrically speaking, a homogeneous polynomial equation of degree 2, say $aX^2+bY^2=dT^2$, with rational coefficients $a, b, c$, represents a conic $C$ in the projective plane $\mathbf P ^2 (\mathbf Q)$, and $C$ admits a rational point $M_0$ iff $C$ is isomorphic to the projective line $\mathbf P ^1 (\mathbf Q)$. Such an isomorphism can be geometrically realized by making a line $D$ pivot around $M_0$ and taking $C\cap D$. This gives the desired parametrization of the solutions $(X, Y, T)$