---
id: 5900f3ac1000cf542c50febf
challengeType: 5
title: 'Problem 64: Odd period square roots'
---
## Description
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?
## Instructions
## Tests
```yml
tests:
- text: euler64() should return 1322.
testString: assert.strictEqual(euler64(), 1322, 'euler64() should return 1322.');
```
## Challenge Seed
```js
function euler64() {
// Good luck!
return true;
}
euler64();
```
## Solution
```js
// solution required
```