# Proving/Disproving: There is a natural #,x, such that 2^(2^x)+1 is not prime

I'm taking a discrete math course, and we just started proof writing. I'm struggling a little bit with getting the concepts to click when it comes to proving expressions with exponents.

Here's the homework question I'm stuck on. We are either supposed to put "F" if it can be proven to be false, or give an example of the smallest value for which it's true, if it can be proven true.

∃n∈N(¬Prime(2^(2^n) + 1))

I negated the problem and turned it into this:

∀n∈N(Prime(2^(2^n) + 1))

Then I reasoned something like this:

Given a prime number x, x is the product of two positive integers k , j ∈ N By definition of a prime number, either k = 1 and j = x, or k = x and j = 1.

Because 2^n is an integer (side note: I know this to be true, but I don't know by what theorem or proof, and I think I should list it, whatever it is), let m = 2^n.

Thus, x = (2^m) + 1.

Okay, guys. I know how to prove that this is odd, but I don't know how to prove or disprove that it's prime. The only way I know to disprove a prime is to factor a quadratic.

Can anyone guide me towards where to go from here, and/or correct me if I've made a mistake with what I have so far?

Thank you very much in advance

• Have you tried finding such an $n$? – Morgan Rodgers Feb 7 '17 at 2:45
• I don't understand this question. What concept is it trying to test? Anyway, $n=5$, $4294967297 = 671*6700417$ works as a counterexample. – астон вілла олоф мэллбэрг Feb 7 '17 at 2:46
• @астонвіллаолофмэллбэрг.The smaller factor is 641, not 671. (Typo?). – DanielWainfleet Feb 7 '17 at 2:50
• "The only way I know to disprove a prime is to factor a quadratic." There are all sorts of ways to factor a composite integer. For the size of number involved with $n \le 5$ the basic trial division method is fast enough. – hardmath Feb 7 '17 at 2:50
• @user254665 Thank you for pointing out. The actual factorization is $641*6700417$. – астон вілла олоф мэллбэрг Feb 7 '17 at 3:10