<s>[INST] 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 ?
 [/INST] 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.
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