I am not an expert on modulus arithmetic and I am computer scientist looking to see if there is a better way to solve an equation in the form of
$$(x^2 + 43) \mod (44-2x)=0$$
I am currently doing a linear search, I have other equations I am trying to solve other equations that are similar, but the search is becoming inefficient for large values. Any help would be appreciated.
The solutions to this equations are
$$x = -9,5,21,23$$
On
Note that if
$$x^2 + 43 \equiv 0 \pmod{44 - 2x} \tag{1}\label{eq1}$$
then multiplying by $-2$ gives
$$-2x^2 - 86 \equiv 0 \pmod{44 - 2x} \tag{2}\label{eq2}$$
Also, dividing $44 - 2x$ into $-2x^2 - 86$ gives
$$-2x^2 - 86 = \left(x + 22\right) \left(44 - 2x\right) - 1054 \tag{3}\label{eq3}$$
Thus, your original equation can be simplified to just look for cases where
$$-1054 \equiv 0 \pmod{44 - 2x} \tag{4}\label{eq4}$$
As such, you only need to check cases where $44 - 2x$ is a factor of $1054 = 2 * 31 * 17$, in particular, $\pm 1, \pm 2, \pm 31, \pm 17, \pm 62, \pm 34, \pm 527, \pm 1054$. For each value $n$ among these factors, you can then solve for $x$ by using $x = 22 - \frac{n}{2}$. As such, for integral values of $x$, the factor must be even, giving the results of $x = 21, 23, -9, 53, 5, 39, -505, 549$.
Also, if you wish to avoid using negative modulus values, then you need to have $44 - 2x \gt 0$, so $x \lt 22$, giving the results of just $x = -505, -9, 5, 21$.
I assume that you're looking to solve the equation over the integers.
First notice that when $x$ is even, then $x^2+43$ is odd, but the modulus is even. (The modulus is always even.) Therefore in order for $x$ to be a solution, $x$ must be odd. Let $x=2k+1$ then, for some integer $k$. Let's see if that simplifies things.
We now have $$4k^2+4k+44 \equiv 0 \pmod{2(2k-21)}.$$
Cancelling a $2$ from the equation and the modulus at the same time, we now have $$2k^2+2k+22 \equiv 0 \pmod{2k-21},$$ which is much nicer.
The reason it's much nicer is that $2k-21$ has leading coefficient 2, which divides the leading coefficient of $2k^2+2k+22$. Thus, we can subtract the equation $(2k-21)k=2k^2-21k\equiv 0$ to get $$23k + 22 \equiv 0 \pmod{2k-21},$$ and now we can subtract $11(2k-21)= 22k-231\equiv 0$ to get the equation $$k+253\equiv 0 \pmod{2k-21}.$$
Now we rewrite this. We want to find for which $k$ there exist $q$ such that $k+253 = q(2k-21)$.
Rearranging, we have $(2q-1)k = 253+21q$
Thus, $k$ is a solution if and only if $k$ can be written as $\frac{21q+253}{2q-1}=10+\frac{q+263}{2q-1}$.
Thus we have reduced the question to figuring out when $\frac{q+263}{2q-1}$ is an integer.
This is kind of a pain, so let's solve the much easier question of when it is an integer or a half integer, which we can solve by noting that in that case, $$\frac{2(q+263)}{2q-1}=\frac{2q+526}{2q-1}=1+\frac{527}{2q-1}$$ will be an integer. Thus $2q-1$ will be a factor of $527$.
Note that $527$ is odd, so that $527/(2q-1)$ will always be odd, so that the resulting integer will always be even. This means that no value of $q$ gave a half integer solution, so every possible value of $q$ gives a value of $x$. The factors of 527 are $\pm 1$,$\pm 17$,$\pm 31$, and $\pm 527$, corresponding to values for $q$ of $1,0,9,-8,16,-15$, $264$, and $-263$.
These give values of $k$ of $11,10,19,2,26,-5,274$, and $-253$. Thus we get solutions for $x$ of $23,21,39,5,53,-9,549$, and $-505$.