# proof that ${n}^{2}-n+41$ is prime for $0\le n\le 40$

We show that for $0\le n\le 40$, ${n}^{2}-n+41$ 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 (http://planetmath.org/IntegralBinaryQuadraticForms) integral binary quadratic form of discriminant^{} $-163$; that form is ${x}^{2}+xy+41{y}^{2}$. The smallest prime that is represented by that form is $41$. For suppose $p={x}^{2}+xy+41{y}^{2}$ and $$. 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 $n\le 40$. Then

$$ |

and thus ${n}^{2}-n+41$ has a prime factor^{} $$. Write ${n}^{2}-n+41=qc$; then $q{x}^{2}+(2n-1)xy+c{y}^{2}$ represents $q$ ($x=1,y=0$); its discriminant is

$${(2n-1)}^{2}-4qc=4{n}^{2}-4n+1-4({n}^{2}-n+41)=-163$$ |

Since there is only one equivalence class^{} of forms with discriminant $-163$, $q{x}^{2}+(2n-1)xy+c{y}^{2}$ is equivalent to ${x}^{2}+xy+41{y}^{2}$ and thus represents the same integers. But we know that ${x}^{2}+xy+41{y}^{2}$ cannot represent any prime $$, so cannot represent $q$. Contradiction^{}. So ${n}^{2}-n+41$ is prime for $n\le 40$.

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 (http://planetmath.org/IntegralBinaryQuadraticForms) of the appropriate discriminant, which is $1-4p$.

Title | proof that ${n}^{2}-n+41$ is prime for $0\le n\le 40$ |
---|---|

Canonical name | ProofThatN2n41IsPrimeFor0leqNleq40 |

Date of creation | 2013-03-22 16:55:48 |

Last modified on | 2013-03-22 16:55:48 |

Owner | rm50 (10146) |

Last modified by | rm50 (10146) |

Numerical id | 7 |

Author | rm50 (10146) |

Entry type | Proof |

Classification | msc 11A41 |