Re: a 50 $ question on primes

Tim Peters <tim.one@xxxxxxxxxxx> writes:
[marian]
Tried to prove this for two full months and failed ( a friend verified
it on a coputer for triples of primes up to about 12000). It's a
nasty one in my book and it's addictive. Here it is:

Let a,b,c, be distinct primes and let f(X)=X^2-X+1. Then
abc does not divide f(a)f(b)f(c).

Not true for four primes, not true for composites, but in this
incarnation I believe this must be true.

And yes, I'll send the 50$ check to anyone who can prove/disprove it
(to be proudly put on the wall in a nice frame).

How about

a = 349 f(a) = 121453, which is also prime
b = 121453 f(b) = 14750709757 = 7^2 * 31 * 9710803
c = 9710803 f(c) = 94299685194007 = 97 * 349 * 2785563619

Bravo!

FYI, I had no strategy here beyond a "reasonably clever" computer search,
motivated by that "it didn't feel like it should be" true that no such
triple of primes exists. So far, that's the only counterexample the
program has found.

Same here.

? forstep(p=7,7000000,6,if(isprime(p),fp=f(p);ffp=factor(fp)~[1,];for(qi=1,#ffp,if(ffp[qi]>p,fq=f(ffp[qi]);ffq=factor(fq)~[1,];for(ri=1,#ffq,if(f(ffq[ri])%p==0,print(p" "ffp[qi]" "ffq[ri])))))))
349 121453 9710803
[still crunching...]

Phil
--
"Home taping is killing big business profits. We left this side blank
so you can help." -- Dead Kennedys, written upon the B-side of tapes of
/In God We Trust, Inc./.
.

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