<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] This is the partial solution.
$ax^2+by^2-cz^2=d\tag{1}$
Substitute $ x=1, y=pt+1, z=qt+1$ to above equation.
$p,q$ are arbitrary.
If $a+b=c+d$ then we get $t = \large\frac{-2(bp-cq)}{-cq^2+bp^2}.$
If $bp^2-cq^2=\pm1$, we can get the integral solution $(x,y,z).$
Thus, equation $bp^2-cq^2=\pm1$ is reduced to Pell's equation $X^2-bcY^2=\pm b\tag{2}.$
If equation $(2)$ has a solution , then  equation $(1)$ has infinitely many positive integer solutions.
In particular, let $b=1$ then if $c$ is not a perfect square, Pell's equation $X^2-cY^2=1$ always has  infinitely many distinct integer solutions.
Example for $(a,b,c,d)=(4,1,2,3): p^2-2q^2 = 1.$
$(x,y,z)=(1, 7, 5),(1, 239, 169),(1, 8119, 5741),(1, 275807, 195025)$...
Example for $(a,b,c,d)=(14, 1, 13, 2): p^2-13q^2 = 1.$
$(x,y,z)=(1, 2194919, 608761),(1, 3698003921039, 1025641750321)$...
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