The French mathematician Pierre de Fermat (1601-1665) considered the sequence 5, 17, 257, 65537, ... , 2^(2^n) + 1, ... He observed that the first four terms of the sequence (here given) corresponding to n = 1, 2, 3, 4, are primes. (A prime number is a natural number other than one divisible only by itself and one. A natural number is divisible by another natural number if the ratio of numbers is itself a natural number.) He conjectured that the following terms were also primes and, being confident his induction was correct, challenged the prominent English mathematicians of the day to prove his conjecture. (Fermat was unable to do so himself.) Discuss the validity of the conjecture and the efforts placed in proving or disproving this conjecture.
\[2^{2^{5}}+1\] Sadly, Fermat's time period did not allow such a calculation.
\(2^{2^6}+1\) is also not prime.
So far only the first 4 terms that Fermat calculated are prime.
@KingGeorge, I kept remembering there was some really fatal irony here... Now I know what that fatal irony was.
less than 100 years later, Euler found a counter-example. (so not much time spent on this problem as compared to his "Last Theorem" According to http://www.sciencenews.org/view/generic/id/3615/title/Math_Trek__Cracking_Fermat_Numbers people have spent their time looking for divisors of these numbers.
However, there are still many interesting properties to these numbers. Such as if \(m\neq n\) are two positive integers, then \[\large 2^{2^m}+1\]and \[\large 2^{2^n}+1\]share no common divisors.
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You're welcome, Sam! Goodluck.
Y cant i give all of u guys medals. NO FAIR :(
Here's a solution: Give me a medal. I'll give KG a medal. KG will give Phi a medal. WE ALL WIN!
hahah i was thinking abt that lol
(I'm being serious.)
Any objections, @Phi and @KingGeorge ?
kinnnnnggggg medal limitless lol
I've got no objections
and limitless medal phi
Works
oh wait, what's the plan?
and then we r set to go :)
KG, medal me. Then I'll medal Phi.
SUCCESS
Chrome lion approves.
YAY it worked :D
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