$18a$ and $25a$ both integers, then so is $a$ Let $a\in \mathbb{Q}$ such that $18a$ and $25a$ are integers, then we wish to prove that $a$ must be an integer itself. What that means is that $a=\frac{p}{1}$ where $p \in \mathbb{Z}$. What we do know is that we can express the $\gcd(18,25)$ as: 
$$ \gcd(18,25)=18x +25y$$ Now if $x=y=a$, we are done, since:
$$ \gcd(18,25)=18a +25a=43a$$ as the $\gcd$ is always an integer and so is 43, so $a$ is also an integer.
But, how would I generalise this?

Another way to look at this:  $18a$ and $25a$ are integers.  Therefore, so is $25a-18a = 7a$.
Therefore so is $18a-2(7a) = 4a.$
Therefore so is $7a-4a = 3a.$
Therefore so is $4a-3a = a.$
