In the episode Treehouse of Horror V of the long-running TV series The Simpsons,
ID: 3008260 • Letter: I
Question
In the episode Treehouse of Horror V of the long-running TV series The Simpsons, the following counterexample of Fermat's Last Theorem appears:
1782^12 + 1841^12 = 1922^12
While this is not really a solution to the Fermat equation, let's pretend that it is for the purposes
of this exercise.
(a) Reduce equation (1) to a Fermat equation with exponent p = 3. This part is done
(b) Convert the equation from part (a) into Germain's modular form (see Krantz pp. 202-203),
using q = 2p + 1. Confused on this part
(c) Show that the conclusion of Germain's theorem holds for the equation from part (b). Confused
Explanation / Answer
a. Wiles proved Fermat's Last Theorem. If his proof is correct, no disproof can be found. As one who is not an expert in the field, I accept that his proof is correct, because hordes of number theorists, group theorists and other assorted mathematical geniusses will have pored over his paper for hours; and because he had to correct an error in it. So it's been through that rigorous process and had to have changes made already.
b. If a disproof could be found that simply, do you not think people would have given up a while ago?
c. The claimed disproof is not, because it's false, as a simple check will show. The identity given (for others' reference) is
1782^12 + 1841^12 = 1922^12
Now, the last digit of 1782^12 will be the same as the last digit of 1922^12, as both end in the same digit (in fact, it is 6). But the last digit of 1841^12 is not zero (in fact, it is 1). So the left-hand side ends in a different digit from the right-hand side. Your identity is not. As for P = NP, well, the jury's still out on that one. Wooster 11:51, 19 July 2005
Use that P(Ac ) 0 in (C2). (C4) If A B, P(B) = P(A) + P(B A) P(A). Proof . Use (C1) for A1 = A and A2 = B A. (C5) P(A B) = P(A) + P(B) P(A B). Proof . Let P(A B) = p1, P(A B) = p12 and P(B A) = p2, and note that A B, A B, and BA are pairwise disjoint. Then P(A) = p1+p12, P(B) = p2+p12, and P(AB) = p1+p2+p12. (C6) P(A1 A2 A3) = P(A1) + P(A2) + P(A3) P(A1 A2) P(A1 A3) P(A2 A3) + P(A1 A2 A3) and more generally