是否存在无穷多个正整数n使n²+n+41为合数且不被41整除？

Great question! Let's work through this to see why the answer is yes—there are infinitely many such positive integers ( n ).

Key Observations #

First, recall that ( S(n) = n^2 + n + 41 ). We know it's composite when ( n=40,41 ) (and those are divisible by 41), but we need composites not divisible by 41. The trick here is to find a prime ( p \neq 41 ) such that ( S(n) \equiv 0 \pmod{p} ) has solutions, then use those solutions to generate infinitely many ( n ) where ( S(n) ) is a multiple of ( p ) (hence composite) and not divisible by 41.

Finding a Suitable Prime ( p ) #

Let's pick ( p=43 ) (a prime not equal to 41). Let's solve the congruence:
[
n^2 + n + 41 \equiv 0 \pmod{43}
]
Simplify the constant term: ( 41 \equiv -2 \pmod{43} ), so the equation becomes:
[
n^2 + n - 2 \equiv 0 \pmod{43}
]
Factor this quadratic:
[
(n+2)(n-1) \equiv 0 \pmod{43}
]
This gives two solutions: ( n \equiv 1 \pmod{43} ) or ( n \equiv 41 \pmod{43} ).

Generating Infinitely Many Valid ( n ) #

Take the solution ( n = 1 + 43k ) where ( k ) is a positive integer. Let's compute ( S(n) ):
[
\begin{align*}
S(1+43k) &= (1+43k)^2 + (1+43k) + 41 \
&= 1 + 86k + 43^2k2 + 1 + 43k + 41 \
&= 43^2k2 + 129k + 43 \
&= 43(43k^2 + 3k + 1)
\end{align*}
]
Clearly, ( S(n) ) is a multiple of 43, so it's composite (since ( 43 < S(n) ) for all ( k \geq 1 )). Now, we need to confirm it's not always divisible by 41:

• ( 43 \equiv 2 \pmod{41} ), so substitute into the term inside the parentheses:
  [
  43k^2 + 3k + 1 \equiv 2k^2 + 3k + 1 \pmod{41}
  ]
  This quadratic in ( k ) only equals 0 modulo 41 when ( k \equiv 20 \pmod{41} ) or ( k \equiv 40 \pmod{41} ) (by factoring ( (2k+1)(k+1) \equiv 0 \pmod{41} )). For all other values of ( k ), the term is not divisible by 41, so ( S(n) ) isn't divisible by 41 either.

Since there are infinitely many ( k ) that don't satisfy ( k \equiv 20 ) or ( 40 \pmod{41} ), there are infinitely many ( n = 1 + 43k ) where ( S(n) ) is composite and not divisible by 41.

Example to Verify #

Take ( k=1 ): ( n=44 ), ( S(44)=44^2 +44 +41=1936+44+41=2021 ). ( 2021=43 \times 47 ), which is composite, and ( 2021 \div 41 = 49.29\ldots ) (not an integer), so it's not divisible by 41. Perfect!

Generalizing This Approach #

This isn't unique to ( p=43 )—there are infinitely many primes ( p \neq 41 ) for which ( -163 ) (the discriminant of ( S(n) )) is a quadratic residue modulo ( p ). For each such prime, we can generate infinitely many ( n ) where ( S(n) ) is a multiple of ( p ) (hence composite) and not divisible by 41 (by avoiding the small set of ( k ) that would make ( S(n) ) divisible by 41).

内容的提问来源于stack exchange，提问作者Pet123