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simply plugging $0,1,2,3$, and $4$ into $F_n=2^{2^n}+1$ and then observing that the results are prime. I can't think of any other way to 'prove' it, but the method I have proposed is a verification and not a proof.

stackdsewew
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    What is missing from a verification to be a proof? Nothing. – Did Aug 03 '16 at 07:08
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    Those are small enough numbers to be verified by brute force. When $n=4$ you may want to use this result to narrow down the set of candidate factors. At least if you use paper and pencil. – Jyrki Lahtonen Aug 03 '16 at 07:16
  • For primality testing you don't need to test $4$ as a divisor, but you do need to check $3$ and $5$. – Henrik supports the community Aug 03 '16 at 07:22
  • If you can plug in all cases in a proof that is valid. If you don't feel comfortable calling it a proof, that's not your fault. It's the text's fault for calling it a proof. – fleablood Aug 03 '16 at 07:28
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    You can narrow the search considerably. if $p$ divides $F_n$ then $2^{2^n}\equiv -1\pmod p$ so $2$ has order $2^{n+1}$ $\pmod p$. That implies that $2^{n+1}$ divides $p-1$. Thus, to try to factor $F_5$ say you only need to consider prime factors of the form $64k+1$. – lulu Aug 03 '16 at 10:45

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Listing all possibilities and showing them to all be prime is a perfectly valid proof, provided you prove all those numbers are prime. But it may not be what the asker is looking for.

it's a hire car baby
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  • So in the case of n=3, I get $F_n=257$. Is there some formal method of proving it is prime? – stackdsewew Aug 03 '16 at 07:13
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    You need only attempt to divide by every integer up to the square root of $257$ and prove it does not divide by any. Clearly it's not even so no need to divide by any multiple of $2$. Also, since its digits don't add up to a multiple of $3$, there's no need to divide by any multiple of 3. Clearly not a muiltiple of $5$ as it doesn't end in a $5$ or $0$ so you only need to show it doesn't divide by 7,11 and 13. 250 doesn't divide by 7, so that's out. 253 is divisible by 11, so 11's out and 13 times 20 is 260 so that's out. Therefore it's prime. – it's a hire car baby Aug 03 '16 at 07:18
  • @anna_xox Testing whether it's divisible by any prime $\leq 16$. There are six of them. That should take less than a minute by hand if you know how to check some of them without actually dividing (simple tricks exist for checking divisibility by $2,3,5,7$ and $11$ that I know of) – Arthur Aug 03 '16 at 07:19
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    Again: Brute force. Realise if it's not prime it will have a factor less than it's square root, and then try all the possibilities. – Henrik supports the community Aug 03 '16 at 07:20
  • 257 is small enough that simply stating "257 is prime" is fine. If the reader has doubt the reader can verify it on his own damn time. – fleablood Aug 03 '16 at 07:31
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    @fleablood my 5 year old daughter has no knowledge of how to verify it in her own time. Maths Stack Exchange is for mathematicians of all levels. – it's a hire car baby Aug 03 '16 at 08:45
  • Does your daughter have any way to verify that if $n = 3$ that $2^{2^n} + 1 = 257$? If we are asked to prove $2^{2^n} + 1$ is prime for n < 4;n natural including 0, it is enough to point out $2^{2^0} + 1 = 3$; $2^{2^1} + 1 = 5;$ and $2^{2^3} + 1 = 257$ and claim they are all prime. We don't need to show that $b^0 = 1$ for $b \ne 0$ so $2^0 = 1$ and $b^1 = b$ for all $b$ so $2^{2^0} = 2$ and because $2 + 1 = 3$ and as $2^{2^0} = 2$ we can conclude $2^{2^0} + 1 = 3$ and 3 is prime because numbe can have a divisor larger than itself and the numbers lower than 3 or 0, and 1 and 2 and 3, and z.. – fleablood Aug 03 '16 at 15:47
  • @fleablood that proves you're good at maths, not that you're right about the reader verifying it in their own damn time. – it's a hire car baby Aug 04 '16 at 06:22
  • A student is asked to prove that $2^{2^n} +1$ is prime for 0 $\le$ n < 4. It suffices for the student to simply list the values for $n = 0, 1,2,3,4.$n=0$ yields $2^{2^0} + 1 = 3$, $n = 1 $ yields $2^{2^1} + 1 = 5$.$n = 2 $ yields $2^{2^2}+1=2^4 + 1 = 17$ and $n= 3$ yields $2^{2^3} + 1 = 2^8+1 = 257$. The student can claim those are all prime numbers. It seem utterly picayune and petty (not to mention snide) and irrelevent to the level of the class, to expect the student to need to prove those are prime. Presumably the audience is on the same and can either accept it or verify for themselves. – fleablood Aug 04 '16 at 06:47
  • @fleablood bowing out. – it's a hire car baby Aug 04 '16 at 06:51