The Fermat numbers (F(n)=2^(2^n)+1) are pairwise relatively prime; in other words, each Fermat number is coprime with all the others. The first observation to be made from this is that *there must be an infinite number of primes*, since every new Fermat number (and there are an infinite number of those anyways) shares no factors with any other. The second observation that I have is, "Wow. These numbers increase fast." Essentially, the number of binary digits doubles in going from one Fermat number to the next. The same is (roughly) true of the number of decimal digits (after F(3)).

**Proof that the Fermat numbers are relatively prime**:

Begin with the first Fermat number, F(0)=3. This just so happens to equal the Mersenne prime 2^2-1=3. Note that the second Fermat number, F(1)=5, is equal to 2^2+1. 3 and 5 are both prime, and are automatically coprime. (x+1)*(x-1)=x^2-1, so 3*5=(2^2-1)*(2^2+1)=2^4-1=15. F(2)=17=2^4+1, so 3*5*17=(2^4-1)*(2^4+1)=2^8-1=255. F(3)=2^8+1=257 (just so the pattern is obvious...). Now for some induction on the hypothesis that "the multiplication of the first k Fermat numbers is equivalent to F(k)-2":

i) Show for k=1 (done above).

ii) Assume for k=m, m an abitrary natural number. So F(0)*F(1)*F(2)*...*F(m-1)=F(m)-2 (by assumption).

iii) Try k=m+1. We expect that F(0)*F(1)*...*F(m-1)*F(m)=F(m+1)-2. Substituting from the assumption, F(0)*F(1)*...*F(m-1)*F(m)=(F(m)-2)*F(m)=(2^(2^m)-1)*(2^(2^m)+1)=2^(2^(m+1))-1=F(m+1)-2, which is what we expected. Therefore the hypothesis is true.

The consequence of the above induction is that **every Fermat number is congruent to 2 mod every smaller Fermat number**. 2 is the only number that could divide both of the numbers x and x+2, but every Fermat number is odd, so all Fermat numbers are relatively prime.