False statementConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passverified 2026-08-03 (gpt-5.6-sol-codex-subscription)

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FALSE: every Fermat number $2^{2^{n}} + 1$ is prime

Statement

Powers are the natural powers of Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ in the commutative monoid $(\mathbb{Z},\cdot,1)$ of $(\mathbb{Z}, \cdot, 1)$ is a commutative monoid whose group of units is $\{1, -1\}$ ; equivalently $u \mid 1$ holds exactly for $u = 1$ and $u = -1$ , and $\iota : \mathbb{N} \to \mathbb{Z}$ is the embedding of The naturals embed in the integers; a numeral $k$ inside $\mathbb{Z}$ means $\iota(k)$ .

False claim: every Fermat number is prime (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ). That is: for every $m \in \mathbb{N}$ which is a power of two — meaning $\iota(m) = 2^{k}$ for some $k \in \mathbb{N}$ — the integer

$2^{m} + 1$

is prime.

Witness: $k = 5$ , so $m = 32$ and $\iota(32) = 2^{5}$ . The integer $2^{32} + 1$ is not prime, because

$641 \;\mid\; 2^{32} + 1 ,$

while $641$ is neither $1$ nor $2^{32}+1$ .

Euler's verification is used below rather than a ten-digit division: $641$ is simultaneously $5 \cdot 2^{7} + 1$ and $2^{4} + 5^{4}$ , and those two readings together force $641$ to divide $2^{32}+1$ . Congruence notation is not available at this point in the library, so every step is written as a divisibility statement with an explicit witness.

Facts & Assumptions

Given: The integers $2$ , $5$ , $641$ and the powers named below.

[L1]

$p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ; an integer $> 1$ that is not prime is composite (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ).

[L2]

$g^{0} = 1$ and $g^{\sigma(t)} = g^{t} g$ in $(\mathbb{Z},\cdot,1)$ (Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ , Semigroup and monoid, $(\mathbb{Z}, \cdot, 1)$ is a commutative monoid whose group of units is $\{1, -1\}$ ; equivalently $u \mid 1$ holds exactly for $u = 1$ and $u = -1$ ).

[L3]

Exponent laws for natural exponents in a monoid: $g^{a+b} = g^{a}g^{b}$ , $(g^{a})^{b} = g^{ab}$ , and $(gh)^{b} = g^{b}h^{b}$ when $gh = hg$ (Exponent laws in a group: $g^{m+n} = g^{m}g^{n}$ and $(g^{m})^{n} = g^{mn}$ for all $m, n \in \mathbb{Z}$ , and $(gh)^{n} = g^{n}h^{n}$ when $g$ and $h$ commute).

[L4]

$d \mid a$ means $a = dc$ for some $c \in \mathbb{Z}$ ; divisibility is reflexive and linear, so $d \mid a$ and $d \mid b$ give $d \mid a - b$ (Divisibility in $\mathbb{Z}$ : $d \mid a$ when $a = dq$ for some integer $q$ , Divisibility is reflexive and transitive on $\mathbb{Z}$ , and is linear: if $d \mid a$ and $d \mid b$ then $d \mid ax + by$ for all integers $x, y$ ; also $d \mid a$ implies $d \mid ac$ , $-d \mid a$ and $d \mid -a$ ).

[L6]

$\mathbb{Z}$ is a commutative ring: addition and multiplication are associative and commutative, $x \cdot 1 = x$ , multiplication distributes over addition, and every $x$ has an additive inverse; we write $u - v$ for $u + (-v)$ (The integers form a commutative ring, Arithmetic on the integers, The integers as equivalence classes of pairs of naturals).

[L7]

The order on $\mathbb{Z}$ is total, antisymmetric and transitive, is compatible with addition, and positives are closed under multiplication (The integers form a totally ordered ring, Order on the integers); $\iota$ is injective, preserves addition, multiplication and order, and has as image the nonnegative integers, with $\iota(0) = 0$ , $\iota(1) = 1$ (The naturals embed in the integers, Discreteness: $\sigma(n)$ is the immediate successor, The natural numbers $\mathbb{N}$ (von Neumann), Order on the natural numbers, Addition of natural numbers).

[L8]

Induction on $\mathbb{N}$ (The principle of mathematical induction).

Refutation

technique · direct

1.1

$0 < 1$ , and every integer $y > 0$ satisfies $y \ge 1$ : $y = \iota(t)$ with $t \ne 0$ , so $1 \le y$ because $\iota$ preserves the order.

1.2

The small powers, by repeated doubling from [L2]: $2^{1} = 2$ , $2^{2} = 4$ , $2^{3} = 8$ , $2^{4} = 16$ , $2^{5} = 32$ , $2^{6} = 64$ , $2^{7} = 128$ , $2^{8} = 256$ , $2^{9} = 512$ , $2^{10} = 1024$ . Also $5^{2} = 25$ and $5^{4} = 25 \cdot 25 = 625$ .

L2L3L6algebra

2.1

$2^{t} \ge 1$ for every $t \in \mathbb{N}$ : the set of such $t$ contains $0$ , since $2^{0} = 1$ , and if $2^{t} \ge 1$ then $2^{\sigma(t)} = 2^{t} \cdot 2$ has both factors $\ge 1 > 0$ , so the product is positive and hence $\ge 1$ by step 1.1. Induction finishes it.

step 1.1L2L7L8

2.2

$641 = 5 \cdot 2^{7} + 1$ and $641 = 2^{4} + 5^{4}$ : indeed $5 \cdot 128 = 640$ and $640 + 1 = 641$ , while $16 + 625 = 641$ .

step 1.2L6algebra

3.1

Put $u := 5 \cdot 2^{7}$ , so $u = 640$ and $u + 1 = 641$ ; hence $641 \mid u + 1$ .

step 2.2L4

3.2

$2^{32} + 1 > 641$ : by [L3], $2^{32} = 2^{10} \cdot 2^{22} = 1024 \cdot 2^{22}$ , and $2^{22} \ge 1 > 0$ , so $2^{32} \ge 1024 > 641$ . Hence $641 \ne 2^{32}+1$ ; also $641 \ne 1$ and $641 > 0$ , and $2^{32}+1 > 1$ .

step 2.1step 1.2L3L7algebra

4.1

$u^{4} - 1 = (u+1)(u^{3} - u^{2} + u - 1)$ , by expanding the right side with distributivity: the terms $-u^{3} + u^{3}$ , $u^{2} - u^{2}$ and $-u + u$ cancel. Hence $641 \mid u^{4} - 1$ .

step 3.1L2L4L6

5.1

$u^{4} = (5 \cdot 2^{7})^{4} = 5^{4} (2^{7})^{4} = 5^{4} \cdot 2^{28}$ , using [L3] with $5$ and $2^{7}$ commuting and $7 \cdot 4 = 28$ . So $641 \mid 5^{4} \cdot 2^{28} - 1$ .

step 4.1L3L6

6.1

From step 2.2, $5^{4} = 641 - 2^{4}$ , so $5^{4} \cdot 2^{28} - 1 = (641 - 2^{4}) 2^{28} - 1 = 641 \cdot 2^{28} - 2^{4} 2^{28} - 1 = 641 \cdot 2^{28} - (2^{32} + 1)$ , using $2^{4} 2^{28} = 2^{32}$ from [L3].

step 2.2step 5.1L3L6

7.1

Now $641 \mid 641 \cdot 2^{28}$ , and $641 \mid 641 \cdot 2^{28} - (2^{32}+1)$ by steps 5.1 and 6.1; subtracting, linearity gives $641 \mid 2^{32} + 1$ .

step 5.1step 6.1L4L6

8.1

So $2^{32}+1$ has a positive divisor other than $1$ and itself, hence is not prime, and being greater than $1$ it is composite. Since $\iota(32) = 2^{5}$ by step 1.2, the natural number $32$ is a power of two and the claim fails there.

step 1.2step 7.1step 3.2L1L7∎

Remarks

The first five Fermat numbers are prime, which is why the claim was believed. For $k = 0,1,2,3,4$ the values are $3$ , $5$ , $17$ , $257$ and $65537$ , all classically known to be prime; that is not verified here, since the refutation needs only the single witness at $k = 5$ . Fermat conjectured primality for all $k$ ; Euler produced the divisor $641$ .
Why the two readings of $641$ are both needed. $641 = 5 \cdot 2^{7} + 1$ is what makes $641$ divide $5^{4}2^{28} - 1$ , and $641 = 2^{4} + 5^{4}$ is what converts that into a statement about $2^{32}+1$ . Either alone says nothing.
Nothing is claimed about the cofactor. The refutation needs only that $641$ is a divisor of $2^{32}+1$ lying strictly between $1$ and $2^{32}+1$ ; the complementary factor is neither computed nor analysed here, and its primality is not asserted.

Depends on

Used by

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Sources

Fermat number (Wikipedia) (standard reference, not scraped)
Millersville University notes: Fermat numbers (standard reference, not scraped)

FALSE: every Fermat number 22n+12^{2^{n}} + 122n+1 is prime