How to solve a diophantine equation $\, j\, a + k\, b = n\, ab\,$ for $a,b$?

Question

I was preparing for the Maclaurin Olympiad and I came across this question:

Show that: $$ \frac{1}{a} + \frac{1}{b} = \frac{5}{11}$$ has no solutions for positive integers $a,b$.

Is there any general method to solve equations like this:

$$11a + 11b = 5ab$$

Thank you

Answer 1

One way: rewrite as $$ (5a - 11)(5b-11) = 11^2 $$ and consider all ways of factoring the right side.

Answer 2

Now express one of them, say $b$: $$b(5a-11)=11a$$ so $$5a-11\mid 11a$$

so $$5a-11 \mid 5\cdot 11a-11(5a-11) =121$$

so $$5a-11\in \{-1,1,-11,11,-121,121\}$$

so $$5a\in \{10,13,0,22,-110,132\}$$ so $a=2$ or $a=0$ or $a=-22$. Clealy, noone fits.

Answer 3

You can just eliminate possibilities.

• Both must be greater than $2$ as otherwise the sum exceeds $\frac12>\frac5 {11}$

• One must be a multiple of $11$ since $5ab=11(a+b)$ and $11$ is prime. So the other must be no bigger than $\frac{11}{4}$ since $\frac{1}{\frac{11}{4}}+\frac{1}{11} = \frac{5}{11}$

• But $3 =\frac{12}{4}>\frac{11}{4}$, so there are no possibilities

Answer 4

If $a=b$, so $$\frac{2}{a}=\frac{5}{11},$$ which is impossible.

Let $a>b$.

Thus, $$\frac{5}{11}<\frac{2}{b}$$ or $$b<\frac{22}{5},$$ which gives $$b\leq4.$$ In another hand, $$\frac{1}{b}<\frac{5}{11}$$ or $$b>\frac{11}{5},$$ which gives $$b\geq3.$$ Now, easy to check that for $b\in\{3,4\}$ we can not get a natural value of $a$.