3

As you know, $x^2+x+41$ is composite when $x=41$ but, how can I claim that $41$ is the smallest positive integer satisfying condition ? Please explain why

J.U.math
  • 1,351
  • 1
    This is answered in https://math.stackexchange.com/questions/53686/to-what-extent-can-values-of-n-such-that-n2-n41-is-composite-be-predicted?rq=1 . – Patrick Stevens Aug 08 '16 at 18:18
  • 1
    Then, can i deduce n^2+n+41 in the same way ? – J.U.math Aug 08 '16 at 18:20
  • The smallest positive integer satisfying which condition? $x^2+x+m$ is always composite when $x=m$, for any positive integer $m$. – Robert Israel Aug 08 '16 at 18:32
  • See this answer. Theorem $\ $ The polynomial $\rm\ f(x)\ =\ (x-\alpha):(x-\alpha')\ =\ x^2 + x + k\ $ assumes only prime values for $\rm\ 0\ \le\ x\ \le\ k-2 \ \iff\ \mathbb Z[\alpha]\ $ is a PID. – Bill Dubuque Aug 08 '16 at 18:42
  • 1
    You can't claim that $41$ is the smallest such positive integer, because it's not. If $x = 40$, then $x^2 + x + 1 = 40^2 + 40 + 1 = 40(40 + 1) + 41 = 40\cdot 41 + 41 = 41(40 + 1) = 41^2$, which is composite.. – Dylan Aug 08 '16 at 18:43

1 Answers1

5

The discriminant of $x^2+x+41$ is $-163$ and the class number of $\mathbb{Q}(\sqrt{-163})$ is one (see Stark-Heegner's theorem), hence it follows that the first few values of $x^2+x+41$ are all primes. On a smaller scale, the same happens for $x^2-x+11$ or $x^2-x+17$, for instance (since both $\mathbb{Q}(\sqrt{-43})$ and $\mathbb{Q}(\sqrt{-67})$ have class number one). But you may just check by hand that if $f(x)=x^2+x+41$ then $f(0),f(1),\ldots,f(39)$ are all prime numbers.

Jack D'Aurizio
  • 353,855
  • 1
    There is no need to check by hand since one know precisely which "first few values" are prime by an old result of Rabinowitsch (see here), which yileds a very nice conceptual explanation of the essence of the matter. – Bill Dubuque Aug 08 '16 at 18:48
  • 1
    $f(40)$ is actually composite (contrary to OP's desired claim in the question) but otherwise, checking manually is probably the best way to go for those without the number theory background. This is especially true for such an easy polynomial and even more so if a computer is available. –  Aug 08 '16 at 18:53