2018-10-10 22:03:03 +00:00
---
id: 5900f3ac1000cf542c50febf
2021-02-06 04:42:36 +00:00
title: 'Problem 64: Odd period square roots'
2018-10-10 22:03:03 +00:00
challengeType: 5
2021-02-06 04:42:36 +00:00
forumTopicId: 302176
2021-01-13 02:31:00 +00:00
dashedName: problem-64-odd-period-square-roots
2018-10-10 22:03:03 +00:00
---
2020-12-16 07:37:30 +00:00
# --description--
2018-10-10 22:03:03 +00:00
2021-02-06 04:42:36 +00:00
All square roots are periodic when written as continued fractions and can be written in the form:
2018-10-10 22:03:03 +00:00
2021-02-06 04:42:36 +00:00
$\\displaystyle \\quad \\quad \\sqrt{N}=a_0+\\frac 1 {a_1+\\frac 1 {a_2+ \\frac 1 {a3+ \\dots}}}$
2018-10-10 22:03:03 +00:00
2021-02-06 04:42:36 +00:00
For example, let us consider $\\sqrt{23}:$:
2018-10-10 22:03:03 +00:00
2021-02-06 04:42:36 +00:00
$\\quad \\quad \\sqrt{23}=4+\\sqrt{23}-4=4+\\frac 1 {\\frac 1 {\\sqrt{23}-4}}=4+\\frac 1 {1+\\frac{\\sqrt{23}-3}7}$
2018-10-10 22:03:03 +00:00
2021-02-06 04:42:36 +00:00
If we continue we would get the following expansion:
2018-10-10 22:03:03 +00:00
2021-02-06 04:42:36 +00:00
$\\displaystyle \\quad \\quad \\sqrt{23}=4+\\frac 1 {1+\\frac 1 {3+ \\frac 1 {1+\\frac 1 {8+ \\dots}}}}$
2018-10-10 22:03:03 +00:00
2021-02-06 04:42:36 +00:00
The process can be summarized as follows:
2018-10-10 22:03:03 +00:00
2021-02-06 04:42:36 +00:00
$\\quad \\quad a_0=4, \\frac 1 {\\sqrt{23}-4}=\\frac {\\sqrt{23}+4} 7=1+\\frac {\\sqrt{23}-3} 7$
2018-10-10 22:03:03 +00:00
2021-02-06 04:42:36 +00:00
$\\quad \\quad a_1=1, \\frac 7 {\\sqrt{23}-3}=\\frac {7(\\sqrt{23}+3)} {14}=3+\\frac {\\sqrt{23}-3} 2$
2018-10-10 22:03:03 +00:00
2021-02-06 04:42:36 +00:00
$\\quad \\quad a_2=3, \\frac 2 {\\sqrt{23}-3}=\\frac {2(\\sqrt{23}+3)} {14}=1+\\frac {\\sqrt{23}-4} 7$
2020-12-16 07:37:30 +00:00
2021-02-06 04:42:36 +00:00
$\\quad \\quad a_3=1, \\frac 7 {\\sqrt{23}-4}=\\frac {7(\\sqrt{23}+4)} 7=8+\\sqrt{23}-4$
2020-12-16 07:37:30 +00:00
2021-02-06 04:42:36 +00:00
$\\quad \\quad a_4=8, \\frac 1 {\\sqrt{23}-4}=\\frac {\\sqrt{23}+4} 7=1+\\frac {\\sqrt{23}-3} 7$
2020-12-16 07:37:30 +00:00
2021-02-06 04:42:36 +00:00
$\\quad \\quad a_5=1, \\frac 7 {\\sqrt{23}-3}=\\frac {7 (\\sqrt{23}+3)} {14}=3+\\frac {\\sqrt{23}-3} 2$
2020-12-16 07:37:30 +00:00
2021-02-06 04:42:36 +00:00
$\\quad \\quad a_6=3, \\frac 2 {\\sqrt{23}-3}=\\frac {2(\\sqrt{23}+3)} {14}=1+\\frac {\\sqrt{23}-4} 7$
2020-12-16 07:37:30 +00:00
2021-02-06 04:42:36 +00:00
$\\quad \\quad a_7=1, \\frac 7 {\\sqrt{23}-4}=\\frac {7(\\sqrt{23}+4)} {7}=8+\\sqrt{23}-4$
2018-10-10 22:03:03 +00:00
2021-02-06 04:42:36 +00:00
It can be seen that the sequence is repeating. For conciseness, we use the notation $\\sqrt{23}=\[4;(1,3,1,8)]$, to indicate that the block (1,3,1,8) repeats indefinitely.
2018-10-10 22:03:03 +00:00
2021-02-06 04:42:36 +00:00
The first ten continued fraction representations of (irrational) square roots are:
2018-10-10 22:03:03 +00:00
2021-02-06 04:42:36 +00:00
$\\quad \\quad \\sqrt{2}=\[1;(2)]$, period = 1
2018-10-10 22:03:03 +00:00
2021-02-06 04:42:36 +00:00
$\\quad \\quad \\sqrt{3}=\[1;(1,2)]$, period = 2
2018-10-10 22:03:03 +00:00
2021-02-06 04:42:36 +00:00
$\\quad \\quad \\sqrt{5}=\[2;(4)]$, period = 1
2020-12-16 07:37:30 +00:00
2021-02-06 04:42:36 +00:00
$\\quad \\quad \\sqrt{6}=\[2;(2,4)]$, period = 2
2020-12-16 07:37:30 +00:00
2021-02-06 04:42:36 +00:00
$\\quad \\quad \\sqrt{7}=\[2;(1,1,1,4)]$, period = 4
2020-12-16 07:37:30 +00:00
2021-02-06 04:42:36 +00:00
$\\quad \\quad \\sqrt{8}=\[2;(1,4)]$, period = 2
2020-12-16 07:37:30 +00:00
2021-02-06 04:42:36 +00:00
$\\quad \\quad \\sqrt{10}=\[3;(6)]$, period = 1
2020-12-16 07:37:30 +00:00
2021-02-06 04:42:36 +00:00
$\\quad \\quad \\sqrt{11}=\[3;(3,6)]$, period = 2
2020-12-16 07:37:30 +00:00
2021-02-06 04:42:36 +00:00
$\\quad \\quad \\sqrt{12}=\[3;(2,6)]$, period = 2
2020-12-16 07:37:30 +00:00
2021-02-06 04:42:36 +00:00
$\\quad \\quad \\sqrt{13}=\[3;(1,1,1,1,6)]$, period = 5
Exactly four continued fractions, for $N \\le 13$, have an odd period.
How many continued fractions for $N \\le 10\\,000$ have an odd period?
2020-12-16 07:37:30 +00:00
# --hints--
2021-02-06 04:42:36 +00:00
`oddPeriodSqrts()` should return a number.
```js
assert(typeof oddPeriodSqrts() === 'number');
```
`oddPeriodSqrts()` should return 1322.
2018-10-10 22:03:03 +00:00
```js
2021-02-06 04:42:36 +00:00
assert.strictEqual(oddPeriodSqrts(), 1322);
2018-10-10 22:03:03 +00:00
```
2020-08-13 15:24:35 +00:00
2021-01-13 02:31:00 +00:00
# --seed--
## --seed-contents--
```js
function oddPeriodSqrts() {
return true;
}
oddPeriodSqrts();
```
2020-12-16 07:37:30 +00:00
# --solutions--
2021-01-13 02:31:00 +00:00
```js
// solution required
```