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A Fermat number is the positive integer $F_n=2^{2^n}+1$, indexed by a nonnegative integer $n$. The first five are $3,5,17,257,65537$, all prime numbers; the next is the composite number $F_5=4294967297=641\cdot 6700417$. This family offers a compact lesson in number theory: a very persuasive finite pattern can fail, while an exact divisibility relation settles a gigantic case with little computation.

Why does the exponent itself have the form $2^n$? Suppose $m=ab$ is a positive integer with an odd integer factor $b>1$. The polynomial factorization $x^b+1=(x+1)(x^{b-1}-x^{b-2}+\cdots -x+1)$ shows that $2^a+1$ is a proper divisor of $2^m+1$. Consequently, if $2^m+1$ is a prime number, $m$ must be a power of two. That is a necessary condition, rather than a promise that the surviving candidates are prime.

The Fermat number product identity makes distinct Fermat numbers coprime integers. Thus every term has a prime factor absent from all earlier terms, whether or not that term is itself a prime number. This gives an explicit route to infinitely many different prime numbers, without establishing that infinitely many terms of this particular sequence are prime. Meanwhile the prime divisor of a Fermat number criterion filters possible factors through a small multiplicative order calculation.

The family also links arithmetic to Euclidean geometry: a constructible regular polygon can have a number of sides equal to a power of two times a product of distinct Fermat primes. The Gauss-Wantzel theorem gives the precise criterion. For instance, $15=3\cdot 5$ and $17$ qualify, but $9=3^2$ does not. The word "distinct" matters. Carl Pomerance describes this connection in Primality testing: then and now by Carl Pomerance (slide 18).

Historically, Leonhard Euler records Pierre de Fermat's proposed primality pattern, then refutes it by finding the factor $641$ of $F_5$. His paper was presented in 1732 and published in 1738; those are different events, not competing dates for one publication.^[ref]

A short modern check needs only two representations of $641$:In modular arithmetic these give $5^4{2}^{28}\equiv 1$ and $5^4\equiv -2^4(\mathrm{mod}641)$. Substitution yields $2^{32}\equiv -1(\mathrm{mod}641)$, so $641\mid F_5$. This verifies the factor directly; it does not claim to reconstruct Euler's search process.

A Fermat prime is a Fermat number that is also a prime number:The defining form alone does not imply primality. For example, $65537$ is a Fermat prime, while $4294967297$ is a Fermat number with proper divisors. For $k\ge 1$, every Fermat prime is one modulo four and two modulo three. Pépin's test gives an exact primality testing criterion for these candidates, rather than relying on their form.

For $N\ge 3$, a regular polygon with $N$ sides is a constructible regular polygon exactly when $N=2^a{p}_1\cdots p_r$, where $a$ is a nonnegative integer and the $p_j$ are distinct Fermat primes; the product may be empty. Thus seventeen and fifteen qualify, while nine does not. Carl Friedrich Gauss established sufficiency, and Pierre Wantzel proved necessity in 1837.^[ref]

Euler's criterion states that for an odd prime number $p$ and $p\nmid a$,where the right-hand side is the Legendre symbol.

Let $g$ be a primitive root modulo $p$. Its multiplicative order is $p-1$, so $g^{(p-1)/2}\ne 1$. This power has square $g^{p-1}=1$, and the only roots of $X^2-1$ modulo the odd prime $p$ are $\pm 1$. ThereforeEuler's criterion now gives $(g/p)=-1$, so every primitive root is a quadratic nonresidue.

Now let $p=2^{2^k}+1$ be a Fermat prime, with $k\ge 1$. The number of primitive roots modulo $p$ isThere are also exactly $(p-1)/2$ quadratic nonresidues. Since every primitive root is a nonresidue, these equally large sets coincide. Thus every quadratic nonresidue modulo $p$ is a primitive root, as summarized by Quadratic nonresidues modulo a Fermat prime.

Finally, $p\equiv 1(\mathrm{mod}4)$ and $p\equiv 2(\mathrm{mod}3)$. Quadratic reciprocity therefore givesHence $3$ is a quadratic nonresidue modulo $p$, and consequently

The characteristic of a field is the least positive integer $p$ with $p{1}_F=0$, or zero if none exists. Finiteness forces a positive characteristic, since the sequence of integer multiples of one must repeat. If this least integer factored into two smaller positive factors, their nonzero field multiples would multiply to zero, contradicting the absence of zero divisors. Hence it is prime. The field is a finite-dimensional vector space over its prime subfield ${\mathbb{F}}_p$, so choosing a basis gives $|F|=p^a$.

A nonzero polynomial of degree $n$ over a field has at most $n$ roots: each root supplies a linear factor by polynomial division, and induction applies to the quotient. In particular $X^n-1$ has at most $n$ roots. The elements of $F^{\times }$ whose orders divide $n$ are precisely those roots. The permitted finite-group criterion therefore proves $F^{\times }$ is cyclic.

For the final assertion take an odd Fermat prime $p=2^m+1$, $m\ge 1$. Primality forces $m$ to be a power of two: an odd factor $s>1$ in $m=2^{r}s$ would factor $(2^{2^r}{)}^s+1$ nontrivially. Since $2^m\equiv -1(\mathrm{mod}p)$, the order of two divides $2m$ but not $m$. As $2m$ is a power of two, its only divisor not dividing $m$ is $2m$ itself. ThusIt generates the full group exactly when $2m=2^m$, which holds for $m=1,2$ and fails for $m\ge 3$. Therefore the possibilities are $p=3,5$, equivalently $p\le 5$ among odd Fermat primes. The oddness is implicit in referring to two as a unit; $p=2$ is not an admissible case for that claim.

The characteristic of a field is the least positive integer $p$ with $p{1}_F=0$, or zero if none exists. Finiteness forces a positive characteristic, since the sequence of integer multiples of one must repeat. If this least integer factored into two smaller positive factors, their nonzero field multiples would multiply to zero, contradicting the absence of zero divisors. Hence it is prime. The field is a finite-dimensional vector space over its prime subfield ${\mathbb{F}}_p$, so choosing a basis gives $|F|=p^a$.

A nonzero polynomial of degree $n$ over a field has at most $n$ roots: each root supplies a linear factor by polynomial division, and induction applies to the quotient. In particular $X^n-1$ has at most $n$ roots. The elements of $F^{\times }$ whose orders divide $n$ are precisely those roots. The permitted finite-group criterion therefore proves $F^{\times }$ is cyclic.

For the final assertion take an odd Fermat prime $p=2^m+1$, $m\ge 1$. Primality forces $m$ to be a power of two: an odd factor $s>1$ in $m=2^{r}s$ would factor $(2^{2^r}{)}^s+1$ nontrivially. Since $2^m\equiv -1(\mathrm{mod}p)$, the order of two divides $2m$ but not $m$. As $2m$ is a power of two, its only divisor not dividing $m$ is $2m$ itself. ThusIt generates the full group exactly when $2m=2^m$, which holds for $m=1,2$ and fails for $m\ge 3$. Therefore the possibilities are $p=3,5$, equivalently $p\le 5$ among odd Fermat primes. The oddness is implicit in referring to two as a unit; $p=2$ is not an admissible case for that claim.