--- In

primenumbers@yahoogroups.com,

Phil Carmody <thefatphil@...> asked:

> Why's everyone only replying off-list?

Perhaps because the mods have set that as the on-line default?

I tried to reply, but maybe I forgot that trap.

There is no way of telling, since the default

reply is not even copied to oneself.

Here goes again (sorry if it eventually appears twice):

> 3a) Can you categorise p,q,r such that Phi(pqr) only has

> coefficients in {-1,0,1}?

> b) Can you find a bigger pqr with that property than

> anyone else?

pqr=

103534127524074104115541098969958708089926902711923091490097\

001545810093892338679248846099390229911806167096020172161339\

248099921117956336949559455817976090558652472327237037907245\

079673380295270552412505821551314360478060083799787164050405\

497609391864910896620579324752947949306589953853514558460087\

480667962113549945569489945478510973601172041512307421653813\

8676796204064794079723241321918976844647423

has a cyclotomic polynomial that is easily proven to be flat,

once you have factorized pqr into its 3 constituent primes.

Yet I believe that I am the only person in the world

who can actually prove that Phi(pqr) is indeed flat,

if the CIA has not yet spooked my hard drive.

Proof:

print(concat(isprime([p,q,r]),[pqr==p*q*r,r>q,q>p,Mod(r,q*p)^2==1]))

[1, 1, 1, 1, 1, 1, 1]

David (with thanks to Nathan)