Editing Fermat number (section)

==Other theorems about Fermat numbers==
{{math theorem|name=Lemma.|If ''n'' is a positive integer,
::<math>a^n-b^n=(a-b)\sum_{k=0}^{n-1} a^kb^{n-1-k}.</math>
{{math proof|
<math>\begin{align}
(a-b)\sum_{k=0}^{n-1}a^kb^{n-1-k} &=\sum_{k=0}^{n-1}a^{k+1}b^{n-1-k}-\sum_{k=0}^{n-1}a^kb^{n-k}\\
&=a^n+\sum_{k=1}^{n-1}a^kb^{n-k}-\sum_{k=1}^{n-1}a^kb^{n-k}-b^n\\
&=a^n-b^n
\end{align}</math>
}}
}}
{{math theorem| If <math>2^k+1</math> is an odd prime, then <math>k</math> is a power of 2.
{{math proof|If <math>k</math> is a positive integer but not a power of 2, it must have an odd prime factor <math>s > 2</math>, and we may write <math>k= rs</math> where <math>1 \le r < k</math>.

By the preceding lemma, for positive integer <math>m</math>,

:<math>(a-b) \mid (a^m-b^m)</math>

where <math> \mid </math> means "evenly divides". Substituting <math>a = 2^r, b = -1</math>, and <math>m = s</math> and using that <math> s </math> is odd,

:<math> (2^r+1) \mid (2^{rs}+1), </math>

and thus

:<math> (2^r+1) \mid (2^k+1). </math>

Because <math>1 < 2^r+1 < 2^k+1</math>, it follows that <math>2^k+1</math> is not prime. Therefore, by [[contraposition]] <math>k</math> must be a power of 2.
}}}}
{{math theorem| A Fermat prime cannot be a [[Wieferich prime]].
{{math proof| We show if <math>p=2^m+1</math> is a Fermat prime (and hence by the above, ''m'' is a power of 2), then the congruence <math>2^{p-1} \equiv 1 \bmod {p^2}</math> does not hold.

Since <math>2m |p-1</math> we may write <math>p-1=2m\lambda</math>. If the given congruence holds, then <math>p^2|2^{2m\lambda}-1</math>, and therefore

:<math>0 \equiv \frac{2^{2m\lambda}-1}{2^m+1}=(2^m-1)\left(1+2^{2m}+2^{4m}+\cdots +2^{2(\lambda-1)m} \right) \equiv -2\lambda \pmod {2^m+1}.</math>

Hence <math>2^m+1|2\lambda</math>, and therefore <math>2\lambda \geq 2^m+1</math>. This leads to <math>p-1 \geq m(2^m+1)</math>, which is impossible since <math>m \geq 2</math>.
}}}}
{{math theorem|note=[[Édouard Lucas]]| Any prime divisor ''p'' of <math>F_n = 2^{2^n}+1</math> is of the form <math>k2^{n+2}+1</math> whenever {{nowrap|''n'' > 1}}.
{{math proof|title=''Sketch of proof''|Let ''G''<sub>''p''</sub> denote the [[Multiplicative group of integers modulo n|group of non-zero integers modulo ''p'' under multiplication]], which has order {{nowrap|''p'' − 1}}. Notice that 2 (strictly speaking, its image modulo ''p'') has multiplicative order equal to <math>2^{n+1}</math> in ''G''<sub>''p''</sub> (since <math> 2^{2^{n+1}}</math> is the square of <math>2^{2^n}</math> which is −1 modulo ''F''<sub>''n''</sub>), so that, by [[Lagrange's theorem (group theory)|Lagrange's theorem]], {{nowrap|''p'' − 1}} is divisible by <math>2^{n+1} </math> and ''p'' has the form <math>k2^{n+1} + 1</math> for some integer ''k'', as [[Euler]] knew. Édouard Lucas went further. Since {{nowrap|''n'' > 1}}, the prime ''p'' above is congruent to 1 modulo 8. Hence (as was known to [[Carl Friedrich Gauss]]), 2 is a [[quadratic residue]] modulo ''p'', that is, there is integer ''a'' such that <math>p|a^2-2.</math> Then the image of ''a'' has order <math>2^{n+2}</math> in the group ''G''<sub>''p''</sub> and (using Lagrange's theorem again), {{nowrap|''p'' − 1}} is divisible by <math>2^{n+2}</math> and ''p'' has the form <math>s2^{n+2} + 1</math> for some integer ''s''.

In fact, it can be seen directly that 2 is a quadratic residue modulo ''p'', since

:<math>\left(1 +2^{2^{n-1}} \right)^{2} \equiv  2^{1+2^{n-1}} \pmod p.</math>

Since an odd power of 2 is a quadratic residue modulo ''p'', so is 2 itself.
}}}}

A Fermat number cannot be a perfect number or part of a pair of [[amicable numbers]]. {{harv|Luca|2000}}

The series of reciprocals of all prime divisors of Fermat numbers is [[Convergent series|convergent]]. {{harv|Křížek|Luca|Somer|2002}}

If {{nowrap|''n''<sup>''n''</sup> + 1}} is prime and <math>n \ge 2</math>, there exists an integer ''m'' such that {{nowrap|1=''n'' = 2<sup>2<sup>''m''</sup></sup>}}. The equation
{{nowrap|1=''n''<sup>''n''</sup> + 1 = ''F''<sub>(2<sup>''m''</sup>+''m'')</sub>}}
holds in that case.<ref>Jeppe Stig Nielsen, [http://jeppesn.dk/nton.html "S(n) = n^n + 1"].</ref><ref>{{MathWorld|urlname=SierpinskiNumberoftheFirstKind|title=Sierpiński Number of the First Kind}}</ref>

Let the largest prime factor of the Fermat number ''F''<sub>''n''</sub> be ''P''(''F''<sub>''n''</sub>). Then,
:<math>P(F_n) \ge 2^{n+2}(4n+9) + 1.</math> {{harv|Grytczuk|Luca|Wójtowicz|2001}}