Positive Prime Factor?

Question

Let a є 4N+3 = {4n+3 : n є N}. Show that a has a positive prime factor in 4N+3. (N is the set of all natural numbers)

I'm not quite sure how to go about this... any help would be really appreciated!

Answer 1

What you're asking is to find all n,m ε N such that
4n+3 is prime and 4m+3 = 0(mod(4n+3)).

Note that, in general, for a given m there will not exist a corresponding n. In particular, when 4m+3 is, itself, prime no such n will exist. Fortunately the question does not require that extra condition so a single example suffices to show that there is at least one such pair n,m.

One such pair is 1,8 since 4*1+3=7, 4*8+3=35 and 7 is a prime factor of 35.

There *is* a general solution for finding n,m pairs, but I'll leave the fun of finding it to you. (Hint: Remember how to do modular arithmetic ☺)


Doug

Answer 2

Suppose the contrary, then a is can be factored as a set of p's all of which are = 4m+1. Then, of course, a would be of the variety 4k+1, contradiction.

Steve

Answer 3

a={0,7,11,15,19,23,...}

15=3*5
3 is prime

7,11,19,23 are primes