$x^2 + x + 41$ is composite [duplicate]

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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

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1 Answer

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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.

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