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I could calculate the following $$2+1=3$$ $$2^2+1=5$$ $$2^{2^2}+1=17$$ $$2^{2{^{2^2}}}+1=65537$$

Now how can I prove or disprove the formula always gives a prime number

E.H.E
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  • Very little is known about the prime-ness of $2^{2^n}+1$ in general, and these are all of this form. It is not even known that there are infinitely many primes of this more limited form - indeed, the only such "Fermat primes" currently known are $n=0,1,2,3,4.$ So you'll have a hard time getting much help on this site proving the above. https://en.wikipedia.org/wiki/Fermat_number – Thomas Andrews Dec 12 '15 at 23:38
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    @E.H.E Come on! You can't be this naive. –  Dec 12 '15 at 23:50
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    Also, this appears to be a duplicate of your own question. – Bamboo Dec 12 '15 at 23:52

2 Answers2

3

According to the OEIS (here) the fifth term in this sequence,

$$2^{65536} + 1$$

is divisible by 825753601. Therefore, not all such numbers are prime.

The OEIS link above provides some references on this sequence. As Thomas Andrews states in his comment, this is a subsequence of the Fermat numbers, and not much is known.

Bamboo
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2

Entering the following into Sage:

(2^2^2^2^2+1).is_prime()

yields

False

(Also, WolframAlpha says "unknown" when asked that question.)

Therefore, the formula does not always give a prime number.

Eric M. Schmidt
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