The Problem:
Find integers $m$ and $n$ such that $14m+13n=7$.
Where I Am:
I understand how to do this problem when the number on the RHS is $1$, and I understand how to get solutions for $m$ and $n$ in terms of some arbitrary integer through modular arithmetic, like so:
$$14m-7 \equiv 0 \pmod {13} \iff 14m \equiv 7 \pmod {13}$$
$$\iff m \equiv 7 \pmod {13} $$
$$\iff m=7+13k \text{, for some integer }k.$$
And repeating the same process for $n$, yielding
$$ n=-7(2k+1) \text{, for some integer } k. $$
I then tried plugging these in to the original equation, thinking that I only have one variable, $k$, to solve for, but they just ended up canceling. The only way I can think to proceed from here is brute force, but I imagine there's a more elegant way to go about this. Any help would be appreciated here.
Best Answer
If $14x+13y=1$ then multiplying by $7$ gives $14(7x)+13(7y)=7.$