False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passverified 2026-08-03 (gpt-5.6-sol-codex-subscription)

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FALSE: $n^{2} + n + 41$ is prime for every natural number $n$

Statement

False claim: for every $n \in N$ the integer

$ι (n)^{2} + ι (n) + 41$

is prime (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ), where $ι : N \to Z$ is the embedding of The naturals embed in the integers and the square is the natural power of Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ in the commutative monoid $(Z, \cdot, 1)$ of $(Z, \cdot, 1)$ is a commutative monoid whose group of units is ${1, - 1}$ ; equivalently $u ∣ 1$ holds exactly for $u = 1$ and $u = - 1$ . As usual a numeral $k$ inside $Z$ means $ι (k)$ .

Witness: $n = 40$ . Here

$40^{2} + 40 + 41 = 1600 + 40 + 41 = 1681 = 41 \cdot 41,$

so $1681$ has the positive divisor $41$ , which is neither $1$ nor $1681$ : it is composite, not prime.

The failure is structural rather than accidental: $40^{2} + 40 = 40 \cdot 41$ , so the whole expression is $40 \cdot 41 + 41 = 41 \cdot 41$ .

Facts & Assumptions

Given: The integers $40$ , $41$ and $1681$ .

[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 $g^{σ (k)} = g^{k} g$ in $(Z, \cdot, 1)$ , so $g^{2} = g \cdot g$ (Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ , Exponent laws in a group: $g^{m + n} = g^{m} g^{n}$ and $(g^{m})^{n} = g^{m n}$ for all $m, n \in Z$ , and $(g h)^{n} = g^{n} h^{n}$ when $g$ and $h$ commute, Semigroup and monoid, $(Z, \cdot, 1)$ is a commutative monoid whose group of units is ${1, - 1}$ ; equivalently $u ∣ 1$ holds exactly for $u = 1$ and $u = - 1$ ).

[L3]

$d ∣ a$ means $a = d c$ for some $c \in Z$ (Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ , Divisibility is reflexive and transitive on $Z$ , and is linear: if $d ∣ a$ and $d ∣ b$ then $d ∣ a x + b y$ for all integers $x, y$ ; also $d ∣ a$ implies $d ∣ a c$ , $- d ∣ a$ and $d ∣ - a$ ).

[L5]

$Z$ is a commutative ring: multiplication is associative and commutative, $x \cdot 1 = x$ , and multiplication distributes over addition (The integers form a commutative ring, Arithmetic on the integers, The integers as equivalence classes of pairs of naturals).

[L6]

The order on $Z$ is total, antisymmetric and transitive and is compatible with addition (The integers form a totally ordered ring, Order on the integers); $ι$ is injective and order preserving with image the nonnegative integers, $ι (0) = 0$ , $ι (1) = 1$ (The naturals embed in the integers, Discreteness: $σ (n)$ is the immediate successor, The natural numbers $N$ (von Neumann), Order on the natural numbers).

Refutation

technique · direct

1.1

$0 < 1$ , since $1 = ι (1)$ is nonnegative and differs from $0 = ι (0)$ by injectivity of $ι$ .

1.2

At $n = 40$ the expression equals $41 \cdot 41$ . By [L2], $40^{2} = 40 \cdot 40$ , and distributivity gives $40 \cdot 40 + 40 = 40 \cdot (40 + 1) = 40 \cdot 41$ ; adding $41$ and using distributivity again, $40 \cdot 41 + 41 = (40 + 1) \cdot 41 = 41 \cdot 41$ .

L2L5

1.3

Arithmetically $40 \cdot 40 = 1600$ , $1600 + 40 = 1640$ , $1640 + 41 = 1681$ , and $41 \cdot 41 = 1681$ ; so the value at $n = 40$ is $1681$ .

L5algebra

2.1

$41 ∣ 1681$ , since $1681 = 41 \cdot 41$ ; and $41 > 0$ .

step 1.2step 1.3L3

2.2

$41 \neq 1$ , because $41 - 1 = 40 > 0$ ; and $41 \neq 1681$ , because $1681 - 41 = 1640 > 0$ ; and $1681 > 1$ .

step 1.1step 1.3L6algebra

3.1

So $1681$ has a positive divisor other than $1$ and itself, hence is not prime; being greater than $1$ it is composite. The claim therefore fails at $n = 40$ .

step 2.1step 2.2L1∎

Remarks

Checking small cases proves nothing here, and that is the point of the item. The values at $n = 0, 1, \dots, 39$ are classically known to be prime — this is Euler's polynomial, and the fact is not verified above, since it is not needed for a refutation. A claim that survives forty consecutive tests and fails at the forty-first is exactly the situation a universal statement has to be proved out of, not tested out of.
The witness is not isolated. $n = 41$ fails for the same structural reason: $41^{2} + 41 + 41 = 41 \cdot (41 + 1 + 1) = 41 \cdot 43 = 1763$ , again divisible by $41$ and again neither $1$ nor itself. What both cases exploit is that $41$ divides the constant term, so it divides the whole value whenever it divides $n^{2} + n$ .
The refuted claim is about $N$ , which contains $0$ . At $n = 0$ the value is $41$ , so the claim is not vacuous at its first index; the witness is needed.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

55 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Formula for primes (Wikipedia) (standard reference, not scraped)
Lucky numbers of Euler (Wikipedia) (standard reference, not scraped)
Purdue University MA 341 Lecture 2 (standard reference, not scraped)

FALSE: n2+n+41n2+n+41 is prime for every natural number nn