finding rational but not integer solutions to $x^2+2y^2=11$

Question

I found an integer solution but was not able to find a formula to use for rational numbers. I attempted to use the chord method but that gets really messy quickly.

Answer 1

Imagine we have a rational solution $(x,y)$ to $x^2 + 2y^2 = 11$. Write $x = a/d$ and $y = b/d$ for some integers $a,b,d$ with $d \neq 0$. Then $x^2 + 2y^2 = 11$ tells us that $a^2/d^2 + 2(b^2/d^2) = 11$, or in other words $a^2 + 2b^2 = 11d^2$.

Conversely, if we have a solution $(a,b,d)$ to this new equation with $d \neq 0$, then $(a/d, b/d)$ is a solution to the original equation! To ensure that this solution is not integral, we need to make sure that $\lvert d \rvert > 1$ and $\gcd(a, b, d) = 1$. And of course, since everything is squared, we need only consider positive values of $a$, $b$, and $d$.

Now the problem is easy enough for us to solve by brute force. We can try to pick $d = 2$, and ask which values $a,b \in \mathbb{Z}$ make $a^2 + 2b^2 = 44$. We need $a$ to be even, so in order to have $\gcd(a,b,d) = 1$ we want $b$ to be odd. We can't have $b = 1$ because $42$ is not a perfect square. We can't have $b = 3$ because $26$ is not a perfect square. We can't have $b \geq 5$ since then $a^2 + 2b^2 \geq 50$ for all $a$. Thus, $d = 2$ doesn't work.

What about $d = 3$? This means we're looking for $a,b \in \mathbb{Z}$ such that $a^2 + 2b^2 = 99$. Reducing the equation modulo $9$, we will need $a^2 + 2b^2 \equiv 0 \pmod{9}$. The quadratic residues modulo $9$ are $0$, $1$, $4$, and $7$, so we are forced into one of the following cases:

• $a^2 \equiv b^2 \equiv 0$
• $a^2 \equiv 1$, $b^2 \equiv 4$
• $a^2 \equiv 4$, $b^2 \equiv 7$
• $a^2 \equiv 7$, $b^2 \equiv 1$

The first case is no good because it forces $\gcd(a,b,d) = 3$. To check the next case, we'll first consider $a = 1$, $b = 2$ (the smallest positive values of $a,b$ for which $a^2 \equiv 1 \mod{9}$ and $b^2 \equiv 4 \pmod{9}$). This doesn't work. Then we look at $a = 1$, $b = 7$ (for the same value of $a$, this is the next smallest value of $b$ we can pick for which $b^2 \equiv 4 \pmod{9}$). Bingo! We have a solution $(1,7,3)$.

Thus, $(1/3, 7/3)$ is a non-integer rational solution to the original equation.

Answer 2

$x^2+2y^2=11\tag{1}$
$[x,y]=[3,1]$ is a known solution for equation $(1)$.
Substitute $x=3+t, y=1+kt$ to equation $(1)$, then we get
$$t = \frac{-2(3+2k)}{1+2k^2}$$ Then we get a parametric solution.
$$x = (-3+6k^2-4k)/(1+2k^2)$$ $$y = -(-1+2k^2+6k)/(1+2k^2)$$ Thus, we get infinitely many rational solutions.

Example: $k=1..10$. $[x,y]=[1/3, 7/3], [13/9, 19/9], [39/19, 35/19], [7/3, 5/3], [127/51, 79/51], [189/73, 107/73], [263/99, 139/99], [349/129, 175/129], [447/163, 215/163], [557/201, 259/201]$