Odd period square roots

[mattgathu]

Solution Number: 64

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?

Ref

https://projecteuler.net/problem=64