Does the Diophantine equation $x^2 + y^2 = 12x +7$ have any solution? Does the Diophantine equation $x^2 + y^2 = 12x +7$ have any solution?
Could anyone describe to me what to do in this type of nonlinear Diophantine equations? 
 A: Studying the equation modulo $4$, we see that there can be no solutions, as $x^2$ and $y^2$ are both either $0$ or $1$, and can thus never sum to $12x + 7\equiv 3$.
Looking at a few simple modular congruences like this (especially modulo anything that actually appears in the equation, so that terms disappear) is a good way to exclude the existence of solutions in many cases. It can never be used (by itself) to prove the existence of solutions. And even for equations that have no solutions, modular arithmetic isn't always the way to come to that conclusion. But it's a simple tool that gets the job done often enough.
A: Hint: $x^2+y^2=12x+7$ can be written as $(x-6)^2+y^2=43$, and $43\equiv -1\pmod{4}$.
A: $$x^2+y^2=12x+7$$
$$(x-6)^2+y^2=43$$
Since 43 is not sum of two squares the equation does not have any integer solutions.
A: When I first looked at this question I misread the equation as:
$$x^2+y^2=12z+7$$
But in fact that more general Diophantine equation has no solution since $12z+7=4(3z+1)+3 \equiv3\pmod4$ while $x^2$ and $y^2$ must each be congruent to $0$ or $1 \pmod 4$. So in this case it's as easy to prove impossibility for a more general equation as for the particular case (ie $z=x$).
A lesson that can be drawn is that if both sides of a Diophantine equation are subject to modular constraints on the values they can take, then sometimes this is sufficient to prove impossibility, regardless of the variables on each side.
