0
$\begingroup$

I need to find all the primes of the form $2^{2^n} + 5$ for all $n>0$

I know that $2^{2^0} + 5 = 7$ is prime and is the only prime. Now I just need to prove it.

I make the claim that if $n>=1$ then $2^{2^n} + 5$ is congruent to $0 \bmod 3$ for all $n>=1$

I then using induction on n prove it.

Base Case: $2^{2^1} + 5$ is congruent to $0 \bmod 3$

Induction Hypothesis Prove $2^{2^k} + 5$ is congruent to $0 \bmod 3$

So, $2^2$ is congruent to $1 \bmod 3$ then, $2^{2^{(k+1)}}$ is congruent to $1^{(k+1)} \bmod 3$ then, 2^2^(k+1) + 5 is congruent to $1^(k+1) + 5 = 0 \bmod 3$

Would this be an acceptable way to show this is true by induction?

  • 1
    This is hard to read without formatting. [here](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) is a good tutorial on formatting for this site. – 2017-02-12
  • 2
    To your question: since $2^2\equiv 1 \pmod 3$ we have, for $n≥1$, that $2^{2^n}\equiv 1^{2^{n-1}}\equiv 1 \pmod 3$. That's enough to do what you need. – 2017-02-12
  • 0
    @lulu I don't understand what you're doing their by the way. – 2017-02-12
  • 0
    Do you know modular arithmetic? If not, then: The claim is that $2^{2m}$ is always one more than a multiple of $3$. To prove it note that $4=3+1$ and write $2^{2m}=4^m=(3+1)^m$ and then expand via the binomial theorem to see that $1$ is the only term not divisible by $3$. – 2017-02-12
  • 0
    This might sound dumb but what does that have to with proving 2^2^m + 5 mod 3 is 0? EDIT: also just out of curiousity is their anything wrong with my proof? I'm having trouble understanding still how that proves what I need to prove. – 2017-02-12
  • 0
    Well, since $n≥1\implies 2^{2^n}=3k+1$ your expression is $3k+1+5=3k+6$ which is obviously a multiple of $3$. – 2017-02-12
  • 0
    All this is a lot easier to write out using the notation of modular arithmetic. Definitely worth learning! – 2017-02-12
  • 1
    Note: it looks to me like you are writing $a^{b^c}=(a^b)^c$ which is not what one normally means. normally, $a^{b^c}=a^{(b^c)}$. But possibly I am misreading. – 2017-02-12
  • 0
    Maybe that's why you couldn't read my earlier computation. I wrote $2^{2^n}=2^{(2\times 2^{n-1})}=4^{2^{n-1}}\equiv 1 \pmod 3$ – 2017-02-12

1 Answers 1

0

Since you say that $2^{2^0} + 5 = 7$, obviously you are interpreting this as $2^{(2^0)}$, and not $(2^2)^0$ as lulu suspected. This is the standard interpretation, but it leaves you with a problem:

When you say that $2^2 \equiv 1 \mod 3$ implies that $2^{2^{k+1}} \equiv 1^{k+1} \mod 3$, this makes use of the "associativity of exponentiation". Unfortunately, the world is not so nice: exponentiation is not associative. $a^{(b^c)} \ne (a^b)^c$ in general. In fact, $(a^b)^c = a^{bc}$.

The result may be true, but this doesn't prove it. lulu has given a correct proof: $2^n = 2\cdot 2^{n-1}$, so $$2^{2^n} = 2^{(2\cdot 2^{n-1})} = (2^2)^{2^{n-1}} = 4^{2^{n-1}}$$ and since $4 \equiv 1 \mod 3$, we have $2^{2^n} \equiv 1^{2^{n-1}} \equiv 1 \mod 3$. No induction necessary.