# How to find a positive integer solution?

Is this the proper way?

f =
3*((A^3 + B^3) - (A^3 + B^3)^(2/3)*B -(A^3 + B^3)^(1/3)*A^2 + A^2*B) ==
A^3 + B^3 + (A^3 + B^3) + 3*(A + B + (A^3 + B^3)^(1/3))*(AB + A(A^3 + B^3)^(1/3) +
B(A^3 + B^3)^(1/3)) - 3AB(A^3 + B^3)^(1/3)

FindInstance[{f, A > 0, B > 0, C > 0}, {A, B, C}, Integers, 5]

• a and A are different symbols, as are b and B. Next, AB is interpreted as a single vairbale called AB (and you probably want A*B). Finally your symbol (1/3) should be written out as (1/3) just as you wrote out (2/3). Jan 29, 2019 at 15:38
• @bills thanks done
– Dale
Jan 29, 2019 at 15:40
• There are still ABs occurring in f. Anyways, replacing AB by A B, Reduce[{f, A > 0, B > 0}, {A, B}, Integers] returns False, telling me that there are no solutions. Jan 29, 2019 at 16:09
• You can simplify to f = A^3 - 6 B C^2 - 3 B^2 C + B^3 == 3 A (B^2 + 2 B C + C^2 + 2 A C) && A^3 + B^3 == C^3. Jan 29, 2019 at 20:52

f = A^3 - 6 B C^2 - 3 B^2 C + B^3 == 3 A (B^2 + 2 B C + C^2 + 2 A C) && A^3 + B^3 == C^3

f[A_, B_] := With[{c = (A^3 + B^3)^(1/3)},

and Plot3D[ f[x, y], {x, 0, 2}, {y, 0, 2}] makes it obvious that f[x, y] < 0 for all positive values of x and y.