The idea behind proving any positive integer can be expressed as a difference of certain other numbers. I came up with a question about the idea stated in the title. To be specific I was asked to assess if it is true or not that any positive integer can be expressed as a difference of some multiple of $7$ and some multiple of $6$. I have begun to look for an insight into what would an answer even look like. I have found some information that is has something to do with the prime factors of the numbers in question although by looking at couple of other examples of similar questions I didn't really grasp the idea behind it.
How do you even write this question mathematically?
$$7x-6y = ???$$
I'm choosing "$x$" and "$y$" as different variables since the multiple of $7$ or $6$ that we're choosing don't really have to be the same number in given instance, do they? How would you describe a positive integer on the right side of this equation though? And what do the prime factors of $7$ and $6$ have to do with this problem?
Also - for similar problems I have seen people using the same variable (eg. $7n-6n$) next to whatever numbers they were trying to prove this question for during their analysis. Why is that so?
I would appreciate it a lot if someone introduced me to the general idea of solving problems such as this.
 A: Another way to look at this is "what is the set $H$  all possible integer combinations of $7$ and $6$"? Then to see whether $k$ is an integer combination, you just have to check whether $k$ is in the set $H$.
That seems silly until you realize that $H$ may be easy to describe rather than enumerate. For instance, if I said "$H$ is all even numbers", that makes testing easy.
Let's work only with integers from now on, OK?
Suppose I have $7k + 6p$. I claim I can rewrite that as a combination not of $7$ and $6$, but a combination of $7$ and $6-7$:
\begin{align}
7k + 6p 
&= 7k + 7p + 6p - 7p\\
&= 7(k - p) + (6-7)p
\end{align}
And I could go the other way, too: any combination of $7$ and $6-7$ can also be written as a combination of $7$ and $6$.
If we say that $B(r, s)$ is the set of all combinations of $r$ and $s$, then I've just shown you that
$$
B(7,6) = B(7, 6-7) = B(7, -1).
$$
Now a combination of $7$ and $-1$ is also a combination of $7$ and $1$:
\begin{align}
7k + (-1)p 
&= 7k + 1(-p)
\end{align}
so now we know that
$$
B(7, 6) = B(7, 1)
$$
But $B(7, 1)$ contains any integer $n$, for
$$
n = 7\cdot 0 + 1 \cdot n
$$
Thus "combinations of 7 and 6" amount to the same thing as "all possible integers".
In general, the combinations of $p$ and $q$ (where $q < p$) are the same as the combinations of $p-q$ and $q$, and you can repeat this process until one of the two numbers becomes a $0$. For instance,
$$
B(14, 4) = B(10, 4) = B(6, 4) = B(2, 4) = B(4, 2) = B(2, 2) = B(0, 2)
$$
which is "all multiples of $2$".
When you repeatedly reduce like this, what number do you end up with when you reach a zero? The other number will be the greatest common divisor of the two starting numbers.
So: if someone says "can you write 38 as a combination of 14 and 4?" the answer is "sure, because $38$ is a multiple of 2, which is the GCD of 14 and 4."
Can you write $61$ as a combination of $21$ and $14$? No, because their GCD is $7$, which doesn't divide into $61$.
