A series of 3 numbers B(1),B(2) and B(3) are defined to be in Arithmetic Progression if B(3) - B(2) = B(2) - B(1).
Without referring to Fermat's Last Theorem, prove that it is not feasible to determine three positive integers in Arithmetic Progression with the nineteenth power of the largest integer being equal to the sum of nineteenth powers of the remaining integers.
Assume there exist positive integers in arithmetic progression a, a+d and a+2d such that
a^19 + (a+d)^19 = (a + 2d)^19
Then, a^19 + a^19 = a^19 (mod d), so a = 0 mod d.... (1)
And, 0 + d^19 = 524288 d^19 (mod a), so d = 0 and/or 524288 = 1 (mod a) ...(2)
If d | a, and a | d then a = d. but a^19 + (a+a)^19 = 524288 a^19 which is not equal to (a+2a)^19=1162261467 a^19.
If 524288 = 1 (mod a) then a | 524287 but 524287 is prime, so a = 524287.
But d | a (by (1)) so d = 524287.
But then a = d which has just been proved impossible.
By exhausting these possibilities we have shown the assumption is false.
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Posted by goFish
on 2006-02-06 16:01:36 |
Proof, I think
