$18a$ 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?
 A: All you know is that there are some $x$ and $y$ with that property, but that doesn't imply that you can take $x=y=a$.
Note that $\gcd(18,25)=1$. Therefore, there are integers $x$ and $y$ such that $18x+25y=1$. But then $a=18xa+25ya\in\mathbb Z$, since $18a,25a,x,y\in\mathbb Z$.
A: You can also prove this by contradiction:
Assume $a=\frac{m}{n}$, where $m,n$ are coprime and $n\gt1$. Then, if $18a=k_1\in\mathbb{Z}$, by the fundamental theorem of arithmetic $$n=2^b3^c.$$ However, assuming $$25a=k_2\in\mathbb{Z},$$ implies $n=5$ or $n=25$, which is obviously a contradiction.
A: 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 $4a-3a = a.$
A: Conceptually $\, \dfrac{m}{18} = a = \dfrac{n}{25}\,$ so it's least denominator divides coprimes $18,25$ so is $1,\,$ so $\,a\in\Bbb Z$
Remark $\ $ This is an additive analog of this (multiplicative) group result 
$$ a^{\large 18} = 1 = a^{\large 25}\,\Rightarrow\, {\rm ord}(a)\mid 18,25\,\Rightarrow\, {\rm ord}(a)=1\,\Rightarrow\, a = 1$$
For further discussion see  denominator ideals and order ideals and unique fractionization.
