Problem 64
Odd period square roots

All square roots are periodic when written as continued fractions and can be written in the form:

 N = a0 + 1 a1 + 1 a2 + 1 a3 + ...

For example, let us consider 23:

 23 = 4 + 23 — 4 = 4 + 1 = 4 + 1 123—4 1 + 23 – 37

If we continue we would get the following expansion:

 23 = 4 + 1 1 + 1 3 + 1 1 + 1 8 + ...

The process can be summarised as follows:

 a0 = 4, 123—4 = 23+47 = 1 + 23—37 a1 = 1, 723—3 = 7(23+3)14 = 3 + 23—32 a2 = 3, 223—3 = 2(23+3)14 = 1 + 23—47 a3 = 1, 723—4 = 7(23+4)7 = 8 + 23—4 a4 = 8, 123—4 = 23+47 = 1 + 23—37 a5 = 1, 723—3 = 7(23+3)14 = 3 + 23—32 a6 = 3, 223—3 = 2(23+3)14 = 1 + 23—47 a7 = 1, 723—4 = 7(23+4)7 = 8 + 23—4

It can be seen that the sequence is repeating. For conciseness, we use the notation 23 = [4;(1,3,1,8)], to indicate that the block (1,3,1,8) repeats indefinitely.

The first ten continued fraction representations of (irrational) square roots are:

2=[1;(2)], period=1
3=[1;(1,2)], period=2
5=[2;(4)], period=1
6=[2;(2,4)], period=2
7=[2;(1,1,1,4)], period=4
8=[2;(1,4)], period=2
10=[3;(6)], period=1
11=[3;(3,6)], period=2
12= [3;(2,6)], period=2
13=[3;(1,1,1,1,6)], period=5

Exactly four continued fractions, for N 13, have an odd period.

How many continued fractions for N 10000 have an odd period?

These problems are part of Project Euler and are licensed under CC BY-NC-SA 2.0 UK