Don't understand this problem: There are only 2 pairs of positive integers $(x,y)$ for which... ..both $\frac{21}{x}$ and $\frac{70}{y}$ are in lowest terms and for which $\frac{21}{x} + \frac{70}{y}$ is an integer. One such pair is $(1,1)$. What is the other such pair? 
This is a Mathematics League problem. The answer sheet says, "...GCD of x and 21 is 1. Therefore x divides y." This is concluded from the fact that the fractions are in lowest terms. But I don't really get how that works because I can think of a bunch of ways for the fractions to be in lowest term and NOT divide each other. Such as $\frac{70}{11}$ and $\frac{21}{4}$. 
Thanks

Because we want solution in its lowest term we need $gcd(x,21) = gcd(y,70) = 1$.
Let $n \in \mathbb{N}$ then we have:
$$\frac {21}{x} + \frac {70}{y} = n$$
$$\frac {21}{x} = n - \frac{70}{y}$$
$$\frac {21}{x} = \frac{ny - 70}{y}$$
Cross-multiply and you'll end up with:
$$21y = x(ny-70)$$
Divide both sides by $y$ and we get:
$$21 = \frac{x(ny - 70)}{y}$$
Becuase the LHS is integer, we have that $y\mid x(ny - 70)$. Let $p$ be a divisor of $y$. We'll prove that if $p$ is divisor of $y$ it can't divide $ny - 70$. Because $p$ divides $y$ then $ny$ is divisible by $p$, in order the whole expression to be divisible by $p$, it has to divide $70$. But from $(y,70) = 1$ we obtain that $(p,70) = 1$, so $p$ and $70$ are coprime numbers, so $p \not\mid ny-70$. This implies that every divisor of $y$ divides $x$, i.e. $y\mid x$
Now divide both sides by $x$. We get:
$$\frac{21y}{x} = ny - 70$$
Obviously the RHS is integer so we have $x\mid 21y$, but $(x,21) = 1$, so this implies $x \mid y$. We've obtained that $x\mid y$ and $y\mid x$, this is possible if and only if $x=y$. So the initial equation becomes:
$$\frac {21}{x} + \frac {70}{y} = n$$
$$\frac {91}{x} = n$$
The LHS will be integer if $x$ is a divisor of $91$. We know that divisors of $91$ are $1,7,13$ and $91$. Checking we obtain that for $7$ and $91$ the expression isn't in it's lowest terms so the only solutions are $(x,y) = (13,13), (1,1)$
If $n$ can be negative integer, we need to take in account the negative divisors of $91$ and $-13$ and $-1$ would do the job.
