Quadratic equation with only one solution on integers

Question

$$x^2-y^2=2xy $$ given this Diophantine equation

is also possible to prove or disprove that it has no solution on integers except $ x=y=0$ ?

Can it have rational numbers as a solution ?

Answer 1

I keep trying to show this method, I'll put in extra detail.

If we have ( assume we have) a solution in nonzero integers to $x^2 - 2xy - y^2=0,$ we may force $\gcd(x,y) = 1$ by dividing both variables by the gcd. The new $(x,y)$ are coprime and not both zero. Next, the discriminant of $x^2 - 2xy - y^2$ is $8.$ Then , $8$ is a quadratic residue for any prime $p \equiv 1,7 \pmod 8,$ and a nonresidue for and $q \equiv 3,5 \pmod 8$

We have $$ x^2 - 2xy - y^2 \equiv 0 \pmod 3$$ with $x,y$ coprime and not both zero. From comments, $$ (x-y)^2 \equiv 2 y^2 \pmod 3$$ If we assume $ y \neq 0 \pmod 3$ it has an inverse (mod 3) and $$ \left( \frac{x-y}{y} \right)^2 \equiv 2 \pmod 3 $$ By ordinary trial, we know that there always $u^2 \equiv 0,1 \pmod 3$ so $ \left( \frac{x-y}{y} \right)^2 \equiv 2 \pmod 3 $ is impossible. This contradicts $ y \neq 0 \pmod 3$ so $ y \equiv 0 \pmod 3.$ But then $(x-y)^2 \equiv 0 \pmod 3$ and $x-y \equiv 0 \pmod 3$ so $x \equiv y \pmod 3,$ therefore $x \equiv y \equiv 0 \pmod 3$

THAT IS: both variables are divisible by $3$ so $\gcd(x,y) \neq 1.$ And this contradiction shows that they both must be zero.

The same thing will work for primes $3,5,11,13.$ Try $5,$ you just need to verify that the squares $\pmod 5$ are $0,1,4.$ See Cassels, Rational Quadratic Forms. While this is just an infinite descent argument, I really like to identify the bottom of the descent first, that being coprime variables that are not all zero.

Answer 2

First, consider the parity of $x$ and $y$.

• If one is even and the other is odd, then $x^2 - y^2$ is odd, but $2xy$ is obviously always even. So this can't happen.
• If both are odd, then $x^2 - y^2$ is even, so that's fine. But if we do arithmetic modulo 4 (with $x,y \in \lbrace 1, 3 \rbrace$), then $x^2 - y^2 = 0$, but $2xy = 2$, so these can't be equal.
• But we can't rule out the case that both $x$ and $y$ are both even, as with the trivial solution $x=y=0$.

So $x$ and $y$ must both be even, i.e., there exist integers $m$ and $n$ such that $x=2m$ and $y=2n$.

$$(2m)^2-(2n)^2=2(2m)(2n)$$ $$4m^2 - 4n^2 = 8mn$$ $$m^2 - n^2 = 2mn$$

Now we're back where we started, except with different variables. So by my same parity argument as before, both $m$ and $n$ must themselves be even, and thus $x$ and $y$ are both multiples of 4.

Repeating this reasoning, we can conclude that $x$ and $y$ must also be divisible by 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, and in general any $2^k$ for $k \in \mathbb{N}$. And the only number infinitely divisible by 2 is 0.

Therefore, $x=y=0$, QED.

---

Now, let's consider the rational case, with $x = \frac{p}{q}$ and $y = \frac{r}{s}$ ($p, q, r, s \in \mathbb{Z}$ and $qs \ne 0$).

$$(\frac{p}{q})^2-(\frac{r}{s})^2=2(\frac{p}{q})(\frac{r}{s})$$ $$\frac{p^2}{q^2} - \frac{r^2}{s^2} = \frac{2pr}{qs}$$ $$\frac{p^2s^2 - q^2r^2}{q^2s^2} = \frac{2pr}{qs}$$ $$p^2s^2 - q^2r^2 = 2pqrs$$

Let $u = ps$ and $v = qr$.

$$u^2 - v^2 = 2uv$$

Which is just our familiar no-nonzero-integer-solutions equation with different variable names.

So no, there are no rational solutions either.