proof that is prime for

We show that for $0\leq n\leq 40$ , $n^2-n+41$ is prime. Of course, this can easily be seen by considering the $41$ cases, but the proof given here is illustrative of why the statement is true.

Recall that there is only one reduced integral binary quadratic form of discriminant $-163$ ; that form is $x^2+xy+41y^2$ . The smallest prime that is represented by that form is $41$ . For suppose $p=x^2+xy+41y^2$ and $p<41$ . Then obviously $y=0$ , so $p=x^2$ , which is impossible. Since equivalent forms represent the same set of integers, it follows that any form of discriminant $-163$ represents no primes less than $41$ .

Now suppose $n^2-n+41$ is composite for some $n\leq 40$ . Then $$n^2-n+41\leq 40^2-40+41<41^2$$ and thus $n^2-n+41$ has a prime factor $q<41$ . Write $n^2-n+41=qc$ ; then $qx^2+(2n-1)xy+cy^2$ represents $q$ ($x=1,y=0$ ); its discriminant is $$(2n-1)^2-4qc=4n^2-4n+1-4(n^2-n+41)=-163$$ Since there is only one equivalence class of forms with discriminant $-163$ , $qx^2+(2n-1)xy+cy^2$ is equivalent to $x^2+xy+41y^2$ and thus represents the same integers. But we know that $x^2+xy+41y^2$ cannot represent any prime $<41$ , so cannot represent $q$ . Contradiction. So $n^2-n+41$ is prime for $n\leq 40$ .

This proof works equally well for the other cases mentioned in the parent article, since for each of those cases, there is only one reduced form of the appropriate discriminant, which is $1-4p$ .

proof that is prime for is owned by Roger Lipsett.