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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.

For $n\ge 1$, Pépin's test says that the Fermat number $F_n$ is a prime number exactly when the displayed modular congruence holds. Modular exponentiation evaluates it by exponentiation by squaring without constructing the enormous ordinary power. The index restriction excludes $F_0=3$, where the base itself is divisible by the candidate.^[ref]

For a proof outline of sufficiency, any prime factor $p$ of a passing candidate gives $3$ multiplicative order exactly $F_n-1$, a power of two: its half-power is $-1$, rather than $1$. Fermat's little theorem forces $F_n-1\mid p-1$, so $p\ge F_n$, and hence $p=F_n$. Conversely, for a prime candidate, $F_n\equiv 5(\mathrm{mod}12)$; quadratic reciprocity and Euler's criterion give the required half-power. At $n=1$, the check is $3^2\equiv 4=-1(\mathrm{mod}5)$.