Don't understand this problem: There are only 2 pairs of positive integers $(x,y)$ for which...

Question

..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

Answer 1

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.

Answer 2

Hint: add the numerators then look at the factors of the sum.

Answer 3

I don't understand the hint given in the question. But if $x\neq y$, then multiplying the equation $\frac{21}x+\frac{70}y=k$ by $\min(x,y)$ makes exactly two of the three terms integer which is impossible (suppose for instance $x<y$ then in $\frac{70x}y$ some factors might cancel from $x$ and $y$, but some of $y$ will be left).

Putting $x=y=n$, we must solve $21+70=kn$ with $k\in\Bbb N$ and with $m$ without factors $2,3,7$. Since $21+70=91=7*13$, one has $(k,n)\in\{(91,1),(7,13)\}$ as unique solutions, so $(1,1)$ and $(13,13)$ are the unique pairs for $(x,y)$.

One might have simplified the problem to make computations even simpler. Since $x,y$ cannot have any factor$~7$ but the numerators do, the integer $\frac{21}x+\frac{70}yk$ must be divisible by $7$, and one might as well as for $x,y$ such that the fractions $\frac3x$ and $\frac{10}y$ are on lowest terms and sum to an integer. The argument that $x=y$ is the same as before and it is trivial to see that $k'n=13$ has only two solutions.