<s>[INST] Does the diophantine equation $ax^2+by^2=cz^2+d$ always have solutions? Let $a,b,c,d,x,y,$ and $z \in \mathbb{N}$  where $a,b,c,$ and $d$ are constants but d is allowed to be zero .
$ax^2+by^2=cz^2+d$
First example :
when $a=b=c=1 $ and $d=0$ we have  the equation : $x^2+y^2=z^2$ which I know its general solution .
Second example:
$a=1 ,b=4,c=1$ and $d=0$ we have the equation : $x^2+4y^2=z^2$ . it has solutions.
one solution of it  is: $x=3,y=2,$ and $z=5$
Third example:
$a=2 ,b=3,c=1$ and $d=0$ we have the equation : $2x^2+3y^2=z^2$ .
this example I tried to find solutions among small numbers but I didn't find any solution.
So does $2x^2+3y^2=z^2$ have solutions but I didn't find any ?
or is there proof that $2x^2+3y^2=z^2$ doesn't have any solution?
 [/INST] Well $2x^2+3y^2=z^2$ actually has no solution for positive integers $x,y,z$.
Assume the smallest solution of $(x,y,z)$ (with minimal $z$) is $(r,s,t)$. Consider $\mod 8$, as $2x^2+ 3y^2\equiv 0,2,3,4,5,6 \pmod{8}$ and $z^2 \equiv 0,1,4 \pmod{8}$, we can see $z^2 \equiv 0,4 \pmod{8}$, implies that $z$ is even. Let $z=2p$, then $2x^2+3y^2=4p^2$, which also implies $y$ to be even. Let $y=2q$, then $2x^2+12q^2=4p^2$, which becomes $x^2+6q^2=2p^2$, implying $x$ is also even. Let $x=2r$, then the equation becomes $4r^2+6q^2=2p^2$, which is $2r^2+3q^2=p^2$. Therefore, $(r,q,p)$ is also a solution. However, $p<z$, which means $(r,q,p)$ is a smaller solution, which contradicts the first statement. Therefore, there are no solutions for $(x,y,z)$.
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