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! 

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.
