The following equation has several solutions $2x+3y=73$ Given  $2x+3y=73$ 
so we know that the GCD of these is 1 so i do the Euclidean algorithm
, i got that  $2(3t-73)+3(73+2t)=73$.
now if  x,y is Rational how many solutions?
if  x,y is Integers how many solutions?
if  x,y is Natural how many solutions?
this question made me confused because i thought if i have gcd = c so it means
that the diofant equation has infinite solution. 
thank you guys.
 A: Rewrite the equation to get
$$y=\frac{73-2x}{3}.$$
This shows that for every rational $x$ there is a rational $y$ satisfying the equation. So there are infinitely many solutions in the rational numbers.
This also shows that $y$ is an integer if and only if $x$ is an integer of the form $3n+2$. So there are also infinitely solutions in the integers.
This also shows that if $y$ is a natural number then $73-2x\geq0$ and so $x\leq\frac{73}{2}$. This shows that there are only finitely many solutions in the natural numbers, and it is not hard to list them all.
A: If $2x + 3y = 73$ then $3y = 73 - 2x$ so that $$y = \frac{73 - 2x}{3}.$$ For any rational value of $x$, $$y = \frac{73 - 2x}{3}$$ is a rational number satisfying $2x + 3y = 73$ so there are a countably infinite number of rational solutions.
For integers, note that if $x = -1$ and $y = 1$ then $2x + 3y = 1$. Multiplying by 73, $x= -73$ and $y = 73$ satisfies $2x + 3y = 73$. Finally, for any $n$, $x = -73 + 3n$, $y = 73 - 2n$ satisfy $2x + 3y = 2(-73 + 3n) + 3(73 - 2n) = 73 + 6n - 6n = 73.$
There are a countably infinite number of integer solutions.
