Let $m$ and $n$ be positive integers such that $m(n-m)=-11n+8$ . Find the sum of all possible values of $m-n$.

Question

Let $m$ and $n$ be positive integers such that $m(n-m)=-11n+8$ . Find the sum of all possible values of $m-n$.

after manipulation you get the quadratic $0=m^2-mn+(8-11n)$ from that you get $m=\frac{n \pm\sqrt{n^2-4(8-11n)}}{2}$

but Im not sure how to derive all the solutions from this, hints, suggestions and solutions would all be appreciated

taken from the 2018 BIMC
https://chiuchang.org/imc/wp-content/uploads/sites/2/2018/07/BIMC-2018_Keystage-3_Individual.x17381.pdf

Answer 1

Hint:

It is better to expres $n$ in terms of $m$ and then use divisibility properties $$n = {m^2+8\over m+11} \implies m+11\mid m^2+8$$

Notice $m+11\mid m^2-121$ so $m+11\mid 129=3\cdot 43$ so $m+11\in\{1,3,43,129\}$ so $m=32$ or $m=118$...

Answer 2

Hint: Write your equation in the form $$n=m-11+\frac{129}{m+11}$$

Answer 3

$$m(n-m)=-11n+8\implies n =\frac {m^2+8}{m+11}\implies $$ $$n=m-11+\frac {129}{m+11}\implies $$

$$m-n=11-\frac {129}{m+11}$$ The only integral positive solutions are $$(m,n)=(32,24),(118,108)$$

Thus the desired sum is $18$