For a given $D\in \mathbb{N}$ are there infinite solutions to $D=b^2-4ac$ with $a,b,c \in \mathbb{N}$ I am wondering if for given $D\in \mathbb{N}$ we can find an infinite amount of solutions to $D=b^2-4ac$. Obvious $D$ is the discriminant of a binary quadratic form or a polynomial of degree $2$. Maybe the answer can be given with the use of those, but I do not know. Thanks for your answers.


I am wondering if for given $D\in \mathbb{N}$ we can find an infinite amount of solutions to $D=b^2-4ac$

No, not in general.
Consider $D=1$. Then $b^2-1=4ac$. Hence $(b+1)(b-1)=4ac$.
This obviously has four immediate solutions. $b=\pm 1$ and $a=0\vee c=0$.
Since $4|(b+1)(b-1)$, $b$ has to be odd. Which gives $b=2k+1$ for some $k\in\mathbb{N}$ and we get:
$(2k+2)2k=4k^2+4k=4k(k+1)$
Hence we have an infinte amount of solutions. With $k=a$ and $k+1=c$
Consider $D=3$. Then $3=b^2-4ac$. If $a=c$ we get $3=b^2-4a^2=(b-2a)(b+2a)$ but $3$ is prime. Hence there are only (at most) four solutions, for that case, which result from $b-2a=1 \wedge b+2a=3$ and $b-2a=3\wedge b+2a=1$.
If $a\neq c$.
We get, after adding $1$ to the equation:
$b^2+1=4ac+4=4(ac+1)$
Hence $4|b^2+1$ as above $b$ has to be odd.
$(2k+1)^2+1=4k^2+4k+1+1=4k^2+4k+2=2(2k^2+2k+1)$
$2k^2+2k+1=2(ac+1)$
But $2\nmid 2(k^2+k)+1$
As Lord Shark the Unknwon suggested in the comments, it is enough to view the problem $\mod 4$. Because we can always just factor out the 4 of $D$ and reduce to $0, 1,2, 3$ that way.
The case $D\equiv 0\mod 4$ is trivial and has infinite solutions, since then $b^2=4ac$ and if $b=2k$ we get $4k^2=4ac$. Now $a=c=k$ gives infinite solutions.
The last case is $D\equiv 2\mod 4$. 
This fails for a similar argument like the shown $D=3$ case.
$2=b^2-4ac$.
$b^2=4ac+2=2(2ac+1)$
Yet again, $b$ has to be even. This gives us:
$4k^2=2(2ac+1)\Leftrightarrow 2k^2=2ac+1$
But now the LHS is even, while the RHS is odd.
