all n values that are even will give a non-prime number, as is always even and is always even. even+even gives even numbers bigger than 2, which are never prime. i can't continue from here though, dont know about the odd n values
I got this one from our archives as well.
Show that is not prime for . (n is a positive integer!)
This is a more of a Number Theory problem than algebraic. As my proof relies more on a College Algebra level rather than Number Theory I'm curious to see what you might come up with.
-Dan
all n values that are even will give a non-prime number, as is always even and is always even. even+even gives even numbers bigger than 2, which are never prime. i can't continue from here though, dont know about the odd n values
if n is an odd integer >= 3 and n is not a multiple of 5 then n^4 + 4^n is a multiple of 5 and therefore not a prime
Proof:
n is 1,2,3,or 4 (mod 5)
n^4 is 1 (mod 5)
4^n is -1 (mod 5)
So n^4 + 4^n is congruent to 0 (mod 5)
How do we handle the case n=5, 15, 25, 35, 45, ...... ?
If , then .
If , then , so is a perfect square, and factors as
So, all that is left is to show that for odd , both terms are greater than 1. Since , the first term is guaranteed to be greater than 1. So, we just need to show the second term is greater than one.
Since (for all ), we have
Hey, lots of action on this one! Thanks to all.
SlipEternal gets some extra kudos for checking the "trivial" factorization problem. (That one of the factors might be 1.) I did eventually find the proof online (which was pretty much identical to Slip's and my own) but didn't address the trivial factorization problem.
-Dan