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m=2, n=3 , ratio=2
n=2, m=3, ratio=1

m=1,n=2
m=1,n=3

still missing some of solutions

quite right

I used a method where I assumed m = n + p and drew conclusions from the resulting equations

integer solutions could only exist if p=1, 2, -1 or -2

I'd be interested to know if there is another way of doing this

sure - let me type it up for you...

I then took each case and created a table to see which combinations gave a valid integer ratio

e.g. p=2, n=1, m=n+p=1+2=3, ratio=1

hope there isn't a flaw in my reasoning?

i enjoyed seeing ur solution...

thx :)

do you have an alternative method?

yes its a little bit longer than this.

sure...i'll post it later. :)
Nice to see this group alive again.

thx - and yes - finally it awakes! :)

{m,n,ratio}
{{1, 2, 5}, {1, 3, 5}, {2, 1, 2}, {2, 3, 2}, {3, 1, 1}, {3, 2, 1}}

*point