## Odd period square roots

### Problem 64

Published on Friday, 27th February 2004, 06:00 pm; Solved by 9824

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
| 1 23—4 |  | 1 + | 23 – 3 7

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, |  | 1 23—4 | = | 23+4 7 | = 1 + | 23—3 7
a1 = 1, |  | 7 23—3 | = | 7(23+3) 14 | = 3 + | 23—3 2
a2 = 3, |  | 2 23—3 | = | 2(23+3) 14 | = 1 + | 23—4 7
a3 = 1, |  | 7 23—4 | = | 7(23+4) 7 | = 8 + | 23—4
a4 = 8, |  | 1 23—4 | = | 23+4 7 | = 1 + | 23—3 7
a5 = 1, |  | 7 23—3 | = | 7(23+3) 14 | = 3 + | 23—3 2
a6 = 3, |  | 2 23—3 | = | 2(23+3) 14 | = 1 + | 23—4 7
a7 = 1, |  | 7 23—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?