Prove that if $n,m \in \Bbb N$ then $\frac{1}{n} + \frac{1}{m} = \frac{7}{17} $ doesn't have any solutions. Prove that if $n,m \in \Bbb N$  then $\frac{1}{n} + \frac{1}{m} = \frac{7}{17} $ doesn't have any solutions.
My try
I wrote the equation in $2$ forms, but i can't get anything there without analyzing a lot of cases:
$1º$ $\frac{17}{n} + \frac{17}{m} = 7$ 
$2º$ $17(m+n)=7mn$
In the second case i tried to use the fact that $17$ and $7$ are primes, but i found nothing.
Any hints?
 A: Your second approach is good; you have found that
$$17(m+n)=7mn,$$
and that $7$ and $17$ are primes. This means $17$ divides $mn$, so without loss of generality $17$ divides $m$. In particular $\tfrac{1}{m}\leq\frac{1}{17}$. It follows that
$$\frac{1}{n}=\frac{7}{17}-\frac{1}{m}\geq\frac{6}{17}.$$
Can you take it from here?
A: Proceeding from your equation 2, we get that this is equivalent to
$$ (7m-17) (7n-17) = 17^2. $$
A: Suppose $\tfrac1n+\tfrac1m=\tfrac7{17}$. Clearly $\tfrac1n\leq\tfrac7{17}$ so $n\geq\tfrac{17}{7}>2$. Without loss of generality $\tfrac1n\geq\tfrac1m$, and so
$$\frac2n\geq\frac1n+\frac1m=\frac7{17},$$
which shows that $n\leq\frac{2\times17}{7}<5$. Then $2<n<5$ and plugging in $n=3,4$ into
$$\frac1m=\frac7{17}-\frac1n,$$
shows that there are no positive integer solutions.
A: Hint: Solve for $m$:
$$
m = \frac{17 n}{7 n - 17} = 2 + \frac{3n+34}{7 n - 17}
$$
Then $\dfrac{3n+34}{7 n - 17}$ is an integer and so has absolute value at least $1$. (It clearly cannot be zero.) This reduces the possible $n$ to a small set, and none of them works.
