1/m + 1/n + 1/(m*n^2) = 3/4
or, (4-3m)n^2 + 4mn -4=0
The above equation possesses integer solution in m, if for some integer t :
(4m)^2 +16(4-3m)=t^2
or, (2t+2m-3)(2t-2m+3) = 7
or, (2t+2m-3, 2t-2m+3) = (-1,-7), (-7,-1), (1,7), (7,1)
or, (t,m) = (-2, 3); (-2, 0); (2,0); (2, 3).
Hence, m = 0, 3. But for m =0, the expression 1/m + 1/n - 1(m*n^2) is indeterminate. This is a contradiction.
Accordingly, m=3, so that n = (-12+/-8)/-10 = 2, 2/5; giving:
(m,n) = (3,2) as the only possible solution to the given equation.
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