Re: Conjectures with very large lowest counterexamples
From: Oscar Lanzi III (ol3_at_webtv.net)
Date: 07/06/04
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Date: Mon, 5 Jul 2004 19:18:16 -0500
F_5 is not divisible by 237. The correct divisor is 641 (or the
quoteint obtained when F_5 is divided by 641, which IIRC is also prime).
That Fermat numbers suddenly switch from prime to composite at n = 5 is
not entirely a stroke of luck. Using modular arithmetic (and this is a
good homework exercise for you number theory fans) it can be proven that
if a prime p divides F_n = 2^(2^n)+1, then p = 1 mod 2^(n+1). Thus 641
= 1 mod 2^(5+1) for F_5. When this constraint is applied, it admits no
factors below the square root for F_n with n <= 3, and only the prime
candidates 97 and 193 for F_4 (which doesn't give F_4 much chance of
being composite, so to speak). F_5 is the first Fermat number where so
many factors are allowed by the modular-arithmetic constraint that it's
likely to be composite. If only Fermat had known about this aspect of
his conjecture, surely he would have been a little less rash.
-OL
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