Quadratic equation with only one solution on integers $$x^2-y^2=2xy $$ 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 ?
 A: 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.
A: 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.
