China Mathematical Olympiad

First, we prove a lemma.

Lemma The following Diophantine equation $$x^2+11y^2=4m$$ has a solution $(x_0,y_0)$, such that either $x_0$, $y_0$ are odd numbers or $x_0$, $y_0$ are even numbers with $x_0\equiv (2k+1)y_0(\mathrm{mod}m)$.

Consider the expression $x+(2k+1)y$, where $x$, $y$ are integers, and $0\le x\le 2\sqrt{m}$, $0\le y\le \frac{\sqrt{m}}{2}$.

There are $([2\sqrt{m}]+1)\lfloor \frac{\sqrt{m}}{2}\rfloor +1\ge m$ such expressions. So there exist integers $x_1$, $x_2\in [0,2\sqrt{m}]$, $y_1$, $y_2\in [0,\frac{\sqrt{m}}{2}]$, such that $(x_1,y_1)\ne (x_2,y_2)$, and $$x_1+(2k+1)y_1\equiv x_2+(2k+1)y_2(\mathrm{mod}m),$$ So $$x\equiv (2k+1)y(\mathrm{mod}m),\text{where}x=x_1-x_2,y=y_2-y_1.$$ This means $$x^2\equiv (2k+1{)}^2{y}^2\equiv -11y^2(\mathrm{mod}m),$$ that is, $x^2+11y^2=tm$ for some integer $t$. Since $|x|\le 2\sqrt{m}$, $|y|\le \frac{\sqrt{m}}{2}$, we have $$x^2+11y^2<4m+\frac{11}{4}m<7m.$$ So $1\le t\le 6$. As $m$ is an odd number, obviously the equations $x^2+11y^2=2m$, and $x^2+11y^2=6m$ have no integer solution.

(1) If $x^2+11y^2=m$, then $x_0=2x$, $y_0=2y$ is a solution of $x^2+11y^2=4m$ satisfying the lemma.

(2) If $x^2+11y^2=4m$, then $x_0=x$, $y_0=y$ is a solution of $x^2+11y^2=4m$ satisfying the lemma.

(3) If $x^2+11y^2=3m$, then $(x\pm 11y{)}^2+11(x\mp y{)}^2=9\cdot 4m$. First, assume that $3\mid m$. If $x\not\equiv 0(\mathrm{mod}3)$, $y\not\equiv 0(\mathrm{mod}3)$, and $x\not\equiv y(\mathrm{mod}3)$, then $$x_0=\frac{x-11y}{3},y_0=\frac{x+y}{3}$$ is a solution of $x^2+11y^2=4m$ satisfying the lemma. If $x\equiv y\not\equiv 0(\mathrm{mod}3)$, then $$x_0=\frac{x+11y}{3},y_0=\frac{y-x}{3}$$ is a solution of $x^2+11y^2=4m$ satisfying the lemma. Now suppose $3\mid m$. Then the above are still integer solutions. If $x^2+11y^2=4m$ has an even integer solution $x_0=2x_1$, $y_0=2y_1$, then $$x_1^2+11y_1^2=m\iff 36m=(5x_1\pm 11y_1{)}^2+11(5y_1\mp x_1{)}^2.$$ Since one of $x_1$, $y_1$ is even and the other is odd, so $5x_1\pm 11y_1$, $5y_1\mp x_1$ are odd numbers. If $x_1\equiv y_1(\mathrm{mod}3)$, then $x_0=\frac{5x_1-11y_1}{3}$, $y_0=\frac{5y_1+x_1}{3}$ is a solution of $x^2+11y^2=4m$ satisfying the lemma. If $x_1\not\equiv y_1(\mathrm{mod}3)$, then $x_0=\frac{5x_1+11y_1}{3}$, $y_0=\frac{5y_1-x_1}{3}$ is a solution of $x^2+11y^2=4m$ satisfying the lemma.

(4) If $x^2+11y^2=5m$, then $25\cdot 4m=(3x\mp 11y{)}^2+11(3y\pm x{)}^2$. When $5\nmid m$, if $$x\equiv \pm 1(\mathrm{mod}5),y\equiv \mp 2(\mathrm{mod}5),\text{or }x\equiv \pm 2(\mathrm{mod}5),y\equiv \pm 1(\mathrm{mod}5),$$ then $$x_0=\frac{3x-11y}{5},y_0=\frac{3y+x}{5}$$ is a solution of $x^2+11y^2=4m$ satisfying the lemma. If $x\equiv \pm 1(\mathrm{mod}5)$, $y\equiv \pm 2(\mathrm{mod}5)$, or $x\equiv \pm 2(\mathrm{mod}5)$, $y\equiv \mp 1(\mathrm{mod}5)$, then $$x_0=\frac{3x+11y}{5},y_0=\frac{3y-x}{5}$$ is a solution of $x^2+11y^2=4m$ satisfying the lemma.

When $5\mid m$, then the above are still integer solutions. If $x^2+11y^2=4m$ has an even integer solution $x_0=2x_1$, $y_0=2y_1$, then $$x_1^2+11y_1^2=m,x_1\not\equiv y_1(\mathrm{mod}2),$$ and we have $100m=(x_1\mp 33y_1{)}^2+11(y_1\pm 3x_1{)}^2$. If $x_1\equiv y_1\equiv 0(\mathrm{mod}5)$, or $x_1\equiv \pm 1(\mathrm{mod}5)$, $y_1\equiv \pm 2(\mathrm{mod}5)$, or $$x_1\equiv \pm 2(\mathrm{mod}5),y_1\equiv \mp 1(\mathrm{mod}5),$$ then $x_0=\frac{x_1-33y_1}{5}$, $y_0=\frac{y_1+3x_1}{5}$ is a solution of $x^2+11y^2=4m$ satisfying the lemma. If $x_1\equiv \pm 1(\mathrm{mod}5)$, $y_1\equiv \mp 2(\mathrm{mod}5)$, or $x_1\equiv \pm 2(\mathrm{mod}5)$, $y_1\equiv \pm 1(\mathrm{mod}5)$, then $$x_0=\frac{x_1+33y_1}{5},y_0=\frac{y_1-33x_1}{5}$$ is a solution of $x^2+11y^2=4m$ satisfying the lemma.

The lemma is proved.

From the lemma, if $x^2+11y^2=4m$ has a solution $(x,y)$ with $x$, $y$ being odd numbers, then it has a solution $(x_0,y_0)$ with $x_0$, $y_0$ being even numbers satisfying $x_0\equiv (2k+1)y_0(\mathrm{mod}m)$.

Let $l=2k+1$, the solution of the quadratic equation $$m{x}^2+l{y}_{0}x+n{y}_0^2-1=0$$ is $$x=\frac{-l{y}_0\pm \sqrt{l^2{y}_0^2-4mn{y}_0^2+4m}}{2m}=\frac{-l{y}_0\pm x_0}{2m}.$$ So the equation $$m{x}_1^2+l{y}_0{x}_1+n{y}_0^2-1=0$$ has at least an integer solution $x_1$, i.e. $$m{x}_1^2+l{y}_0{x}_1+n{y}_0^2-1=0.$$ This indicates that $x_1$ is an odd number. Now, from this equation, it follows that $$(2n{y}_0+l{x}_1{)}^2+11x_1^2=4n.$$ This means $x^2+11y^2=4n$ has a solution $(x,y)$ with $x$, $y$ being odd numbers, where $x=2n{y}_0+l{x}_1$, $y=x_1$.

Therefore, at least one of the equations $x^2+11y^2=4m$ or $x^2+11y^2=4n$ has a solution $(x,y)$ with $x$, $y$ odd.