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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.

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    $\begingroup$ Your values statistify for all $k$. $\endgroup$ – wythagoras May 7 '15 at 17:44
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If $14x+13y=1$ then multiplying by $7$ gives $14(7x)+13(7y)=7.$

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    $\begingroup$ Note that for an initial solution of $14x+13y=1$ you can use simply $(x,y)=(1,-1).$ Then the general solution is $x=1+13t,y=-1-14t$ using the usual method of completely solving linear two variable equations. $\endgroup$ – coffeemath May 7 '15 at 18:25
  • $\begingroup$ The previous comment was for the general solution to $14x+13y=1.$ After the multiplication to get $7$ on the right side, the general solution to $14x+13y=7$ becomes $x=7+13u, y=-7-14u.$ $\endgroup$ – coffeemath May 7 '15 at 18:31
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How about, find integers $p,q$ with $14p+13q=1$ and then choose $m=7p, n=7q$, which method shows that the general problem of this kind can be solved.

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Find $m$ and $n$ such that $14m + 13n = 1$ and then multiply them both by $7$.

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  • $\begingroup$ Such that $14m + 13n = 1$? $\endgroup$ – Brian Tung May 7 '15 at 17:49
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If you can solve for $1$, how can you use this to now solve for $7$ ?

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Hint $\ $ The set $S$ of integers of the form $\,14m+13n,\ m,n\in\Bbb Z$ are closed under $\color{#c00}{\rm subtraction}$ and closed under $\color{#0a0}{\rm multiplication}$ by any integer. Thus $\,14,13\in S\,\color{#c00}{\Rightarrow \_\_\in S}$ $\,\color{#0a0}{\Rightarrow\ \_\_ \in S}$

Remark $\ $ This is the key idea behind the use of the Extended Euclidean Algorithm to compute the Bezout identity for the gcd (see here).

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You can use Extended Euclidean Algorithm (EEA) to find integers $x,y$ such that $14x+13y=1$, which exist by Bezout because $(14,13)=1$. I use EEA here as shown in this answer.

$\begin{array}{l|r}&14&14(1)& 13(0)\\\hline &13 & 14(0)& 13(1)\\\hline 14-13(1)& 1&14(1)&13(-1)\end{array}$

So $1\!=\!14(1)\!+\!13(-1)$. Now multiply by $7$ to get $7\!=\!14(7)\!+\!13(-7)$.

It is obvious here without any algorithms that $(x,y)=(1,\!-1)$ works, but it is better to generalize.

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you can find $m_0$ and $n_0$ such that: $$14m+13n=0$$
By using the Extended Euclidean table ($m_0=-13$ and $n_0=14$)

Then, using the same method you can find $m_1$ and $n_1$ such that: $$14m+13n=1$$ Giving ($m_1=1$ and $n_1=-1$)

Finally, you can find the set of solutions to $14m+13n=7$ by selecting:
$$m=7m_1+km_0$$ $$n=7n_1+kn_0$$for some k $\in \Bbb R$

This is all the solutions because the values $7m_1$ and $7n_1$ give the solution to $$14m+13n=7$$ and $m_0$ and $n_0$ give the solution to $$14m+13n=0$$ and the addition of these two will always be $7$

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$14m+13n=7\iff13(m+n)+m=7=13-6\iff m_0=-6$, and $m+n=1\iff$ $n_0=1-m_0=1+6=7$. Now that you have an initial solution, I'm sure that you can find out the others. :-$)$

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