Matematické Fórum

Skrytý text:
As vanok suggested, let's investigate when for some , (i.e. ).
To avoid trivialities we consider . Then we must have so we get an integer on the left hand side; without loss of generality .
Then , where both of the fractions at the right have integer values, so necessarily
and for some , .

a)
In second expression, left hand side must be a multiple of 3, so . However, would imply , . Considering the equation mod 9, we notice this is impossible. That leaves us with .

b)
First expression implies either or . And second option is clearly not compatible with a).

So altogether we get . In that case for .


That implies for any solution of the original problem and we can easily check that the only solutions are and .

... and , as vanok points out. Thanks.