Twobby said:thisisgman said:M+E+H = MEH

You failed algebra didn't you?

Well if we assume M, E and H are all equal to an arbitrary value A, then the equation can be rewritten as a 3A=A^3. Divide out an A, so 3=A^2, therefore A=±root3. We also know intuitively that M, E and H can be 0 for the equation to check out, but now we also know they can equal root3 and -root3.

We know that if two of the variables are equal to 0, then the equation cannot check out as the MEH would always equal to 0, whereas M+E+H would equal a positive or negative integer. However, if one variable is equal to 0, then there are an infinite number of possibilities, if you let those two variables be the same number, but with separate signs.

If M, E and H are not equal to one another, then we can generalize by saying one of the three variable is larger than the other 2 (we'll say H). Therefore M≤E≤H. With this, we know that 3H≥MEH, and therefore 3≥ME. If 3≥ME, M≠0 and E≠0 then (M,E) can only be (±1,±1), (±1,±2), (±1,±3) (M and E values are interchangeable of course).

By substituting our points into M+E+H=MEH, then we can find that there are no possible values of H that satisfy (1,1), (-1,-1), but there are H values that satisfy (1,2), (1,-2), (-1,-2), (-1,2), (1,3), (1,-3), (-1,-3) and (-1,3), which are 3, 1/3, 3/4, -1/3, 1/2, -2, -1/2 respectively.

These are just a few solutions, and technically, thisisgman is infinitely right. He's infinitely wrong too... but still.