Positive integer solutions to the equation $\frac{1}{n}+\frac{1}{m}=\frac{1}{143}$

Question

I need help to calculate the solutions of $\frac{1}{n}+\frac{1}{m}=\frac{1}{143}.$

Thank you, and I'm sorry if this is a duplicate, but I don't see any similar questions.

Answer 1

You have $143(m+n)=mn$. Let $d={\rm gcd}(m,n)$ with $m=dm_1$ and $n=dn_1$; where $(m_1,n_1)=1$. You get $143(m_1+n_1)=dm_1n_1$. From here, you have, by observing $(m_1,m_1+n_1)=(n_1,m_1+n_1)=1$, $m_1n_1 \mid 143$. Recalling $143=11\cdot 13$, the possibilities for the pair $(m_1,n_1)$ are $(1,11),(1,143),(1,13),(11,13)$ (and its permutations/$\pm$ versions). The remaining is a finite casework, that I leave you as an exercise.

Answer 2

Multiply $143mn$ on both sides. $$143m+143n=mn \implies (m-143)(n-143)=143^2=11^2 \cdot 13^2$$ Thus, our solutions would be: $$\{m-143,n-143\}=\{1,20449\},\{11,1859\},\{13,1573\},\{121,169\},\{143,143\},\{169,121\},\{1573,13\},\{1859,11\},\{20449,1\}$$ This can be seen by expressing $143^2$ as products of two integers. Now, you can solve for $m,n$.