Re: Prime factors congruent to 2 (mod 3)

On Jul 24, 7:07 pm, qsymmetry <qsymme...@xxxxxxxxx>
wrote:
On Jul 24, 7:10 pm, "Dr Math" <m...@xxxxxxxxx>
wrote:
LOOK! Another mathforum imbecile.

This remark was hostile an uncalled for.  The
question,
albeit elementary was well posed.  It was a
legitimate
question.

And to answer it, I will give a hint.  A Proof by
contradiction is quite simple.   i.e.

Assume that all prime factors are 1 mod 3.

Thank you for your reply.

 I'd be curious to know why this assumption is
 actually enough? For instance, 3 is  prime, but it
is not
congruent to 2 (mod 3)!

What does this has to do with anything?

There are three types of primes: those that are
congruent to 1 modulo
3; those that are congurent to 2 modulo 3; and 3
(which is congruent
to 0 modulo 3). Just like there are three types of
primes when you
consider their congruence modulo 4 (congruent to 1,
congruent to 3, or
congruent to 2; the latter class containing only the
prime 2).

You have *some* number, n. You are assuming that n is
congruent to 2
modulo 3. Then you are asking about the primes that
may divide n, and
you wonder if there is at least one prime that is
congruent to 2
modulo 3 *and* divides n.

Why would you consider the prime 3? Why bring it up?


Because 3 is a prime congruent to 0 mod 3, so a prime *not* congruent to 2 mod 3, i.e. is one of the types of primes to consider, just as you mentioned.

Wow, and here I thought that was tacitly understood.
.

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