How can you tell if there are integral solutions to to $x$ in $c^2 = ax + b$ where $a$ and $b$ are fixed values and $a,b,c \in \mathbb{Z}$.
I have found that there always exists a solution as long as the constant is a perfect square, $b^2$. The equation can be rewritten as, $$x = \frac{(c-b)(c+b)}{a}$$ We can then let the prime factorization of $a$ be $a_1^{k_1}a_2^{k_2}a_3^{k_3} \cdots a_i^{k_i}$ and split it into two parts $n,m$ with $n = a_1^{k_1}a_2^{k_2} \cdots a_j^{k_j}$ and $m = a_{j+1}^{k_{j+1}}a_{j+2}^{k_{j+2}} \cdots a_i^{k_i}$. Then we can let $c \equiv b \;(\bmod\; n)$ and $c \equiv m-b\;(\bmod\; m)$. This means that $b \in \mathbb{Z}_n$ and $m-b \in \mathbb{Z}_m$. Since $n$ and $m$ are coprime by virtue of being the result of the division of the prime factorization of $a$, then $\langle m \rangle = \mathbb{Z}_n$ and $\langle n \rangle = \mathbb{Z}_m$. Again due to the fact that $n$ and $m$ are coprime, we know that $\mathbb{Z}_n \times \mathbb{Z}_m = \mathbb{Z}_a$ since $a=nm$ and $(m,n)$ generates $\mathbb{Z}_a$. Since $(b,m-b) \in \mathbb{Z}_a$, there exists some $v$ such that $(m,n)^v = (b,m-b)$ which means there is always a solution if the constant is a perfect square.
I don't know how you would try to solve this when the constant is not a perfect square since you would no longer be able to factor the numerator.
I made a program that checks if $b \in [0,999)$ works and for $c \in [0,10000)$ for $a = 10$ and it gives b = $\{1, 4, 5, 6, 9, 10, 11, 14, 15, 16, 19, 20, 21, 24, 25, 26, 29, 30, 31, 34, 35 \cdots\}$ which seems to have a pattern of increasing by $1$, then $3$, then $1$. $11$ has a pattern of $2,1,1,4,2,1$ starting from $1$. Is there an easy way to find these patterns?
I am not sure that I am doing this right, but this is what I see:
b=c^2 (mod a)
For a = 10, we need to find all b such that b = c^2 (mod 10). The squares (quadratic residues) mod 10 are 1,4,5,6, and 9. Your results are those numbers + 10n for some integer n. I don't know why b=1,4 was not in your results.