> B. c^n + d^n > p^n .... (13) > > so p divides c+d, let us write c + d = Mp. > > adding (5) and (6) gives us, a + b = p^n (j+k) + Mp ...... (14) > > Implies p divides a+b, which violates our assumption (5) > > This logic will work even if we assume a = jp^n +/- c and b = kp^n +- d > > > > C. c^n + d^n = 0 possible when c and d are both zero (violates (9)) > > or c = -d and adding (7) and (8) will give us a+b = p^n (j+k) > > and so p divides a+b, which violates our assumption (5) > > > > D. Let us assume p divides(both a+b) and Q. > > We know that the only number that divides a+b and Q is n. > > > > C. Q = n (so n^(n-1) divides a+b) > > Q = a^(n-1) b + a^(n-2)b^2 + .... + b^(n-1) > > Q = n possible if both a and b are equal to 1. > > > > D. Q = n^(n-1) [and n divides a+b] > > then n divides a or b and hence n divides d in (2), violates (1) > > > > C. Q has a factor q (not equal to n) that is not a factor of (a+b). > > This is not possible as shown in (12), (13) and (14) > > > > I think this rounds out the proof.