Can a multiple of $15$ and a multiple of $21$ differ by $1$? I know a solution to this question having to do with the fact that the $\gcd(15, 21) = 3$, so the answer is no.
But I can't figure out what is the reasoning behind this. Any help would be really appreciated! 
 A: A multiple of 15 and a multiple of 21 are both multiples of 3.
The difference between two multiples of 3 is another multiple of 3.
1 is not a multiple of 3.
A: It's really simple, imagine the two numbers as $15n$ and $21k$.
Suppose that it's possible, then: $15n-21k = 1$
As you said $\gcd(15, 21) = 3$, so you can factor out: $3(5n-7k) = 1$.
On the left side, you have a multiple of 3, and in the other you have a 1, which isn't, so you have a contradiction. Note that inverting the subtraction won't change anything.
A: No: you can prove that $\text{gcd}(a, a +b) = \text{gcd}(a,b)$. Therefore, we have that $\text{gcd}(a,a+1) = \text{gcd}(a,1) =1$.
If we now consider multiples of $15$ and $21$, say $k \cdot 15, n \cdot 21$ with $k, n \in \mathbb{Z}$, such that $n \cdot 21 = k \cdot 15 + 1$, then we find that $3$ divides $\text{gcd}(k \cdot 15, n \cdot 21) = \text{gcd}(k \cdot 15, k \cdot 15 + 1) = 1$, which is impossible.
A: Suppose that $15m > 21n$. Then you are asking if:
$$15m - 21n= 1.$$
This corresponds to:
$$15m = 21n + 1.$$
The term on the left is divisible by $3$. What about the term on the right?
Well, $21n$ is divisible by $3$ too. But, if you add $1$ to $21n$, it can't be divisible by $3$ anymore. 
Actually, you have the left term which is divisible by $3$ and the right term which is not divisible by $3$.
For these reasons, you can't find those numbers. 
Similar arguments must be used to prove the same when $21n > 15m$. In this case, you should work on $21n = 15m + 1$.
