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2017-11-08 17:39:50 +01:00
All square roots are periodic when written as continued fractions and can be written in the form:
~~~
sqrt(N) = a0 + 1 / (a1 + 1 / (a2 + 1 / (a3 + ... )))
~~~
For example, let us consider `sqrt(23)`:
~~~
sqrt(23)
= 4 + sqrt(23) - 4
= 4 + 1 / ( 1 / ( sqrt(23) - 4 ) )
= 4 + 1 / ( 1 + ( sqrt(23) - 3 ) / 7)
~~~
If we continue we would get the following expansion:
~~~
sqrt(23) = 4 + 1/(1 + 1/(3 + 1/(1 + 1/(8 + ... ))))
~~~
The process can be summarised as follows:
a | Step 1 | Step 1 |Step 1
-----|------------------------|-------------------------|----------------------
`a0` | `4`, `1/(sqrt(23) - 4` | `1*(sqrt(23) + 4) / 7` | `1 + (sqrt(23) - 3)/7`
`a1` | `1`, `7/(sqrt(23) - 3` | `7*(sqrt(23) + 3) / 14` | `3 + (sqrt(23) - 3)/2`
`a2` | `3`, `2/(sqrt(23) - 3` | `2*(sqrt(23) + 3) / 14` | `1 + (sqrt(23) - 4)/7`
`a3` | `1`, `7/(sqrt(23) - 4` | `7*(sqrt(23) + 4) / 7` | `8 + (sqrt(23) - 4)/1`
`a4` | `8`, `1/(sqrt(23) - 4` | `1*(sqrt(23) + 4) / 7` | `1 + (sqrt(23) - 3)/7`
`a5` | `1`, `7/(sqrt(23) - 3` | `7*(sqrt(23) + 3) / 14` | `3 + (sqrt(23) - 3)/2`
`a6` | `3`, `2/(sqrt(23) - 3` | `2*(sqrt(23) + 3) / 14` | `1 + (sqrt(23) - 4)/7`
`a7` | `1`, `7/(sqrt(23) - 4` | `7*(sqrt(23) + 4) / 7` | `8 + (sqrt(23) - 4)/1`
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.
The first ten continued fraction representations of (irrational) square roots are:
~~~
sqrt( 2) = [1;(2)], period=1
sqrt( 3) = [1;(1,2)], period=2
sqrt( 5) = [2;(4)], period=1
sqrt( 6) = [2;(2,4)], period=2
sqrt( 7) = [2;(1,1,1,4)], period=4
sqrt( 8) = [2;(1,4)], period=2
sqrt(10) = [3;(6)], period=1
sqrt(11) = [3;(3,6)], period=2
sqrt(12) = [3;(2,6)], period=2
sqrt(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?