(*

* Solution to Project Euler problem 58

* Copyright (c) Project Nayuki. All rights reserved.

*

* https://www.nayuki.io/page/project-euler-solutions

* https://github.com/nayuki/Project-Euler-solutions

*)

(*

* From the diagram, let's observe the four corners of an n * n square (where n is odd).

* It's not hard to convince yourself that:

* - The bottom right corner always has the value n^2.

* Working clockwise (backwards):

* - The bottom left corner has the value n^2 - (n - 1).

* - The top left corner has the value n^2 - 2(n - 1).

* - The top right has the value n^2 - 3(n - 1).

*

* Furthermore, the number of elements on the diagonal is 2n - 1.

*)

p = 0;

For[n = 1, True, n += 2,

p += Sum[Boole[PrimeQ[n^2 - k * (n - 1)]], {k, 0, 3}];

If[n > 1 && p / (n * 2 - 1) < 1/10,

Break[]]]

n