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Here is the question from our book:

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Let [itex]F_n = 2^{2^n} + 1[/itex] be the nth Fermat numbers. Use the identity [itex]a^2-b^2 = (a-b)(a+b)[/itex] to show that [itex]F_n - 2 = F_0F_1\cdots F_{n-1}[/itex].

Conclude that [itex](F_n,F_m)=1 \ \forall \ n \neq m[/itex]. Show that this implies the infinitude of the primes.

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The first part is easily done by induction, and using [itex]2^{2^n}=(2^{2_{n-1}})^2[/itex].

It is the second, and third part that I am having trouble with.

I just got an idea for the second. It is easy to show that if [itex]d\mid F_n = 2^{2^n}+1[/itex] and [itex]d\mid F_m = 2^{2^m}+1[/itex] Then we can easily show that d is not even (subtract the two, and we get that d divides an even number, and from the previous things we know d divides an odd number, so d cannot be even.)

Now we can use what we proved in the first part (assume, wlog, n>m) [itex]F_n = (F_m+2)F_mF_{m+1}\cdots F_{n-1} + 2[/itex]

[itex]d\mid F_m[/itex] and [itex]d\mid F_n[/itex] so with a little manipulation and properties we can get[itex]d\mid 2[/itex] and so d=1 or d=2, but d is not even so d=1 (ignoring negative divisors as we are looking at the greatest anyway).

Kind of ugly, but it seems correct.

Any ideas of a more slick proof, or any insight into the problem?

Also, any ideas about how I can get that this implies the infinitude of the primes? (standard proof by contradiction? Let p_1,...p_r, be the list of all Fermat primes. Consider some clever number N, and break up into prime decomp? Break into cases?)

Thanks!

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Let [itex]F_n = 2^{2^n} + 1[/itex] be the nth Fermat numbers. Use the identity [itex]a^2-b^2 = (a-b)(a+b)[/itex] to show that [itex]F_n - 2 = F_0F_1\cdots F_{n-1}[/itex].

Conclude that [itex](F_n,F_m)=1 \ \forall \ n \neq m[/itex]. Show that this implies the infinitude of the primes.

-------

The first part is easily done by induction, and using [itex]2^{2^n}=(2^{2_{n-1}})^2[/itex].

It is the second, and third part that I am having trouble with.

I just got an idea for the second. It is easy to show that if [itex]d\mid F_n = 2^{2^n}+1[/itex] and [itex]d\mid F_m = 2^{2^m}+1[/itex] Then we can easily show that d is not even (subtract the two, and we get that d divides an even number, and from the previous things we know d divides an odd number, so d cannot be even.)

Now we can use what we proved in the first part (assume, wlog, n>m) [itex]F_n = (F_m+2)F_mF_{m+1}\cdots F_{n-1} + 2[/itex]

[itex]d\mid F_m[/itex] and [itex]d\mid F_n[/itex] so with a little manipulation and properties we can get[itex]d\mid 2[/itex] and so d=1 or d=2, but d is not even so d=1 (ignoring negative divisors as we are looking at the greatest anyway).

Kind of ugly, but it seems correct.

Any ideas of a more slick proof, or any insight into the problem?

Also, any ideas about how I can get that this implies the infinitude of the primes? (standard proof by contradiction? Let p_1,...p_r, be the list of all Fermat primes. Consider some clever number N, and break up into prime decomp? Break into cases?)

Thanks!

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