Solve this if you can !!

If we multiply throughout the equation by $10mn$ and by rearranging the terms we get :

$4mn-10n-10m+10=0$ which is equivalent to:

$4mn-10n-10m+25=15$ .Observe that the left side can be factored as followed:

$(2m-5)(2n-5)=15$ .Calculating the negative numbers $15$ can be written in $4$ diffferent ways which are $15=15*1$

$15=3*5$

$15=(-15)(-1)$

$15=(-3)(-5)$ If we suppose $n≥m$ then we will have the pairs $(m;n)=(4;5),(3;10),(0;1),(-5;2)$ ,but the last two couples are not acceptable. If we suppose $n<m$ we will obtain $(m;n)=(5;4),(10;3),(1;0),(2,-5)$ but again the last two couples are not acceptable and therefore we have $\boxed{4}$ couples which are $(m;n)=(3;10),(10;3),(4;5),(5;4)$ .