Stillwell, Elements of Number Theory, Chapter 1 Exercises

Problem 1.1.1 Check that the quadratic function $n^2 + n + 41$ is prime for all small values of $n$ (say, for $n$ up to $30$). \phantomsection \label{Problem 1.1.1}

Solution. Using the matlab package in R, we check whether the quantity $n^2 + n + 41$ is prime for each positive natural number less than or equal to $41$. We find that, for each $n$ less than or equal to $40$, this quantity lacks a proper divisor other than $1$ and is hence prime.

$n$ $n^2 + n + 41$ Prime or Composite
$1$ $43$ Prime
$2$ $47$ Prime
$3$ $53$ Prime
$4$ $61$ Prime
$5$ $71$ Prime
$6$ $83$ Prime
$7$ $97$ Prime
$8$ $113$ Prime
$9$ $131$ Prime
$10$ $151$ Prime
$11$ $173$ Prime
$12$ $197$ Prime
$13$ $223$ Prime
$14$ $251$ Prime
$15$ $281$ Prime
$16$ $313$ Prime
$17$ $347$ Prime
$18$ $383$ Prime
$19$ $421$ Prime
$20$ $461$ Prime
$21$ $503$ Prime
$22$ $547$ Prime
$23$ $593$ Prime
$24$ $641$ Prime
$25$ $691$ Prime
$26$ $743$ Prime
$27$ $797$ Prime
$28$ $853$ Prime
$29$ $911$ Prime
$30$ $971$ Prime
$31$ $1,033$ Prime
$32$ $1,097$ Prime
$33$ $1,163$ Prime
$34$ $1,231$ Prime
$35$ $1,301$ Prime
$36$ $1,373$ Prime
$37$ $1,447$ Prime
$38$ $1,523$ Prime
$39$ $1,601$ Prime
$40$ $1,681$ Composite
$41$ $1,763$ Composite

Problem 1.2. Show nevertheless that $n^2 + n + 41$ is not prime for certain values of $n$.

Solution. Let $n = 41$. Then $$41^2 + 41 + 41 = 41(41 + 1 + 1) = 41 \cdot 43,$$ meaning that $n^2 + n + 41$ has two non-trivial proper divisor and is therefore composite.

Problem 1.3. What is the smallest such value?

Solution. Using the matlab package in R and the table we generated in Problem 1.1.1, we find that $n^2 + n + 41$ is prime provided that $n \leq 39$. But when $n = 40$, we have $40^2 + 40 + 41 = 1681$, which is composite. In particular, it has a proper divisor of $41$ as $1681 = 41^2$. So $n = 40$ is the smallest such value of $n$ for which $n^2 + n + 41$ is not prime.

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