Find the pairs $(m,k)$ solving the diophantine equation $m^2=7k+9$

Question

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 ....

Answer 1

$m^2=7k+9 \Rightarrow m^2=2 \;(\mod 7)$.

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

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

Answer 2

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, $(m.k)=(10,13)$

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

Answer 3

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