IRL_ABooklet_2023

Note that $2023=17^2\cdot 7$ and let $(x,y)$ be a solution. If we assume that $y$ is not divisible by $17$, there exists a number $a$ such that $ay\equiv x(\mathrm{mod}17)$. We then have $$0=x^2-xy+y^2\equiv y^2(a^2-a+1)(\mathrm{mod}17),$$ hence $a^2-a+1\equiv 0(\mathrm{mod}17)$. This implies that $a^6\equiv 1(\mathrm{mod}17)$, because $$a^6-1=(a^3-1)(a^3+1)=(a^3-1)(a+1)(a^2-a+1).$$ On the other hand, by Fermat's little theorem, $a^{16}\equiv 1(\mathrm{mod}17)$. Then $a^2\equiv a^2\cdot a^{16}\equiv a^{18}\equiv (a^6{)}^3\equiv 1(\mathrm{mod}17)$, hence $a\equiv \pm 1(\mathrm{mod}17)$. But for these two values of $a$ we get $a^2-a+1\equiv 2-a\equiv 2\mp 1\not\equiv 0(\mathrm{mod}17)$, a contradiction. Hence, $y$ must be divisible by $17$.

It follows that $x$ must also be divisible by $17$, and we look for solutions $(x,y)=(17u,17v)$. Let $N(x,y)=x^2-xy+y^2$ and note that $$N(x,y)=\frac{(2x-y{)}^2+3y^2}{4}.$$ Then $2023=N(17u,17v)=17^{2}N(u,v)$, iff $7=N(u,v)$, or equivalently $$28=(2u-v{)}^2+3v^2.$$ Because $v^2\le 28/3$ and $28$ is not a square, to find solutions to this equation we only need to consider $v\in \{\pm 1,\pm 2,\pm 3\}$. To a solution with negative $v$ there corresponds one with positive $v$, obtained by changing the signs of both, $u$ and $v$. Because $28-3=5^2$, $28-12=4^2$, and $28-27=1^2$, we have these three cases for solutions with positive $v$ $$v=1,2u-v=\pm 5,v=2,2u-v=\pm 4,v=3,2u-v=\pm 1.$$ The possibilities for solutions $(u,v)$ with positive $v$ are therefore $$(3,1)(-2,1)(3,2)(-1,2)(2,3)(1,3).$$ Thus we obtain $12$ solutions to the original equation, namely $$(x,y)=(17,51),(34,51),(-17,34)$$ and all the variants $(-x,-y)$, $(y,x)$, $(-y,-x)$ of these.