2

Experiment with quadratic expression $ax^2+bx+c$ motivated by Euler's prime generating polynomial $n^2+n+41$

Given that $\color{red}{p(n)=4n^2-4n-1}$

Conjecture: There is always an integer $n$ value range from $\color{blue}{5k+1}$ to $\color{blue}{5k+5}$, for $k\ge0$, that $p(n)$ will always give at least a prime number.

For examples

$k=0$, $n:1-5$ : [$p(2)=7$]

$k=1$, $n:6-10$ :[$p(7)=167$]

$k=2$, $n:11-15$ :[$p(11)=439$]

$k=100$, $n:501-505$:[$p(502)=1006007$]

So far the conjecture seem correct, but we are trying to find an example where it does not give a prime number but seem not working, hence we naturally ask the question

How does one prove or disprove this conjecture?

gymbvghjkgkjkhgfkl
  • 8,090
  • Step 1: Learn programming, so that you can yourself test millions of examples to see whether it holds empirically. But REMEMBER that a million is still infinitely far from infinity. Step 2: Learn more mathematics, so as to mitigate the flaws in Step 1. In particular, you should get an intuitive feel for why certain large classes of conjecture should never be true. (In this case, because primes occur pretty much as randomly as possible, so they should not cling to near any quadratic polynomial, since both get rarer with size. – user21820 Feb 09 '17 at 12:52
  • Best advice ever (+1) – gymbvghjkgkjkhgfkl Feb 09 '17 at 13:25

2 Answers2

5

The conjecture is false; try $k=7\cdot23\cdot 47\cdot 79=597793$. The $5k+1$ version is the only hard one to test, and turns out to be composite (just 2 prime factors though). The others are multiples of $7,23,47,79$ respectively.

Note: The technique I used was to simplify $$p(5k+2)=4(5k+2)^2-4(5k+2)-1=100k^2+60k+7$$ so if $k$ is a multiple of $7$, so is $p(5k+2)$.

vadim123
  • 82,796
3

The conjecture is false. Maple gives the smallest counterexample as $k=36$, when the values are $$130319,\ 131767,\ 133223,\ 134687,\ 136159$$ which are all composite.

David
  • 82,662