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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

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    $\begingroup$ 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. $\endgroup$ – Ryan Greyling Jul 26 at 22:26
  • $\begingroup$ 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 $4$. $\endgroup$ – Ryan Greyling Jul 26 at 22:27
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Your equation of

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

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 $n^2 \equiv 2 \pmod{3}$.

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  • $\begingroup$ For methods of finding $n$, see Wikipedia $\endgroup$ – Robert Israel Jul 26 at 22:50
  • $\begingroup$ @RobertIsrael Thanks for your suggestion. I added the Wikipedia article section link to my answer. $\endgroup$ – John Omielan Jul 26 at 22:56
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$\alpha m + \beta = n^2$

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)

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