## Odd period square roots

### Problem 64

Published on Friday, 27th February 2004, 06:00 pm; Solved by 11402

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 – 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, 1√23—4 = √23+47 = 1 + √23—37 a1 = 1, 7√23—3 = 7(√23+3)14 = 3 + √23—32 a2 = 3, 2√23—3 = 2(√23+3)14 = 1 + √23—47 a3 = 1, 7√23—4 = 7(√23+4)7 = 8 + √23—4 a4 = 8, 1√23—4 = √23+47 = 1 + √23—37 a5 = 1, 7√23—3 = 7(√23+3)14 = 3 + √23—32 a6 = 3, 2√23—3 = 2(√23+3)14 = 1 + √23—47 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?