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Solve

$$m^2=7k+9$$

over the integers

First i rearrange got $m^2-9=7k$ And $(m^2-9)/7=k$ So first $m^2-9$ must be divisible by $7$

So suppose $m=7n , 7n+1 ,...,7n+5$ But it doesn't work ....

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    $\begingroup$ So what you're looking at is $$ m^2 \equiv 2 {\mod 7} $$ I would suggest plugging in some values for $m$ and seeing what happens. $m=1, 2, 3, 4, 5, 6, 7 \ldots$. If you find a solution $m_0$,then $m_0 \pm 7$ is also a solution. Therefore, you only need to look at the first $7$ integers. $\endgroup$ – Matti P. Jan 11 at 12:57
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    $\begingroup$ Hint: $m^2-9=(m-3)(m+3)$, so $7\mid m^2-9\iff7\mid m-3\lor7\mid m+3$, since $7$ is prime. $\endgroup$ – Barry Cipra Jan 11 at 14:57
  • $\begingroup$ If $(m_0,k_0)$ is a solution, then $(m_0+7, k_0+2m_0+7)$ is also a solution $$m_0^2=7k_0+9 \iff m_0^2+14m_0+49=7k_0+9+14m_0+49=7(k_0+2m_0+7)+9 \iff\\ (m_0+7)^2=7(k_0+2m_0+7)+9$$ and start with $(3,0)$. $\endgroup$ – rtybase Jan 11 at 17:49
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$m^2=7k+9 \Rightarrow m^2=2 \;(\mod 7)$.

The only numbers from 1 to 7 satisfying the equation are 3 and 4 ($3^2=9=2; 4^2=16=2$. Then, the solution is of the form:

$m\in\{3+7n, 4+7n, n \in \mathbb{N}\}$; and $k=\frac{m^2-9}{7}$

For instance, if $m=81=4+77$, then $k=936$, and

$$81^2=7 \cdot 936 + 9$$

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Above equation $m^2=7k+9$ has solution's,

$m=7w-4$

$k=(7w-1)(w-1)$, where, 'w' is any integer

For $w=2$ we get, $(m.k)=(10,13)$

$(10)^2=7(13)+9$

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  • $\begingroup$ What about (3,0)? $\endgroup$ – pendermath Jan 11 at 17:27
  • $\begingroup$ It is included $w=1$ $\endgroup$ – Mike Jan 11 at 18:31
  • $\begingroup$ Sorry, I meant (4,1) $\endgroup$ – pendermath Jan 11 at 21:04
  • $\begingroup$ For w=0, you will get (m,k)=(4,1) $\endgroup$ – Sam Jan 11 at 22:10
  • $\begingroup$ If $w=0$, then $m=-4$, not 4, right? $\endgroup$ – pendermath Jan 12 at 22:37
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As you said $7$ must divide $m^2-9=(m-3)(m+3)$. Since $7$ is prime, it must divide either $m-3$ or $m+3$.

Therefore, you have two sets of solutions:

Case 1: $7|m-3$ then $$m-3=7l \\ m=7l+3 \\ 7k+9=m^2=49l^2+42l+9 \\ k=7l^2+6l\\ (k,m)= ( 7l^2+6l, 7l+3)$$

Case 2: $7|m+3$ then $$m+3=7l \\ m=7l-3 \\ 7k+9=m^2=49l^2-42l+9 \\ k=7l^2+6l\\ (k,m)= ( 7l^2-6l, 7l-3)$$

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