# Kind of Diophantine Equation

I was trying to approach an equation of the type:

$$\alpha m + \beta = n^2$$

where $$\alpha$$ and $$\beta$$ are given integer constants and $$m$$ and $$n$$ are integer to be found. Is there a standard way to approach equations like these? If so, can you share me a link or some useful material?

Thanks, Dave

• Here this swc.math.arizona.edu/aws/2006/06CohenLectures.pdf if you're interested, although it might be a bit too comprehensive depending on what you're looking for. – Ryan Greyling Jul 26 at 22:26
• The first thing that jumps out at me is that $\alpha m+\beta$ has to be a square, which implies that $\alpha m+\beta$ is a multiple of $4$ or $1$ greater than a multiple of $4$. – Ryan Greyling Jul 26 at 22:27

$$\alpha m + \beta = n^2 \tag{1}\label{eq1A}$$

means that if $$\alpha = 0$$, there's a solution only if $$\beta$$ is a perfect square, with $$n$$ then being $$\sqrt{\beta}$$ and $$m$$ being any integer.

For $$\alpha \neq 0$$, one other restriction to consider is if there's any prime $$p$$ where the # of factors of it in $$\beta$$, call it $$q$$, is odd and the number of factors of $$p$$ in $$\alpha$$ is $$\gt q$$. In those cases, there are no solutions since $$\alpha m + \beta$$ would have the odd $$r$$ factors of $$p$$, but $$n^2$$ must have an even number of factors of $$p$$.

Apart from the restrictions mentioned above, note you have

$$\beta \equiv n^2 \pmod{\alpha} \tag{2}\label{eq2A}$$

i.e., $$\beta$$ must be a quadratic residue modulo $$\alpha$$. Any $$n$$ which satisfies \eqref{eq2A} will then have a corresponding $$m$$ from \eqref{eq1A} of $$m = \frac{n^2 - \beta}{\alpha}$$. As for finding an $$n$$, as suggested by Robert Israel's comment, the Complexity of finding square roots Wikipedia article section describes several algorithmic methods.

However, note if $$\beta$$ is not a quadratic residue modulo $$\alpha$$, then there are no solutions. For example, if $$\alpha$$ is a multiple of $$3$$ and $$\beta \equiv 2 \pmod{3}$$, there are no solutions since $$2$$ is not a quadratic residue modulo $$3$$, i.e., there's no $$n$$ such that $$n^2 \equiv 2 \pmod{3}$$.

• For methods of finding $n$, see Wikipedia – Robert Israel Jul 26 at 22:50
• @RobertIsrael Thanks for your suggestion. I added the Wikipedia article section link to my answer. – John Omielan Jul 26 at 22:56

$$\alpha m + \beta = n^2$$

Given $$(\alpha,\beta)$$ as,

$$(\alpha,\beta)$$=$$[w^2,4w^2(k^2-1)]$$ then,

w=(2k-3)

$$n=(6k^2-13k+6)$$

$$m=(5k^2-12k+8)$$

For, $$k=3$$, we get:

$$(\alpha,\beta)$$=$$(9,288)$$ and

(n,m)=(21,17)