Question

Given an integer `n` is there an efficient way to find integers `p, q` such that `abs(p^2-n q^2) <= 1` ?

Answer 1

Assuming `n` is not a perfect square:

Find the simple continued fraction expansion of `sqrt(n)`, terminate it just before it repeats and expand to find `p/q`.

Explanation:

To find the simple continued fraction expansion of `sqrt(n)`, use the following algorithm:

`m_0 = 0`
`d_0 = 1`
`a_0 = floor(sqrt(n))`

`m_(i+1) = d_i a_i - m_i`

`d_(i+1) = (n - m_(i+1)^2)/d_i`

`a_(i+1) = floor((a_0 + m_(i+1)) / d_(i+1))`

The continued fraction expansion is then `[a_0; a_1, a_2, a_3,...]`

`sqrt(n) = a_0 + 1/(a_1 + 1/(a_2 + 1/(a_3 + 1/...)))`

This will repeat immediately after the first term satisfying `a_i = 2 a_0`

Stop just before the `a_i = 2 a_0` term and compute the resulting fraction `p/q`. This will satisfy `abs(p^2 - n q^2) = 1`

For example, consider `n = 6798`

`m_0 = 0`, `d_0 = 1`, `a_0 = floor(sqrt(6798)) = 82`

Then:

`m_1 = d_0 a_0 - m_0 = 1*82 - 0 = 82`

`d_1 = (n - m_1^2) / d_0 = (6798 - 82^2) / 1 = 74`

`a_1 = floor((a_0 + m_1) / d_1) = floor((82 + 82) / 74) = 2`

`m_2 = d_1 a_1 - m_1 = 74 * 2 - 82 = 66`

`d_2 = (n - m_2^2) / d_1 = (6798 - 66^2) / 74 = 33`

`a_2 = floor((a_0 + m_2)/d_2) = floor((82 + 66) / 33) = 4`

`m_3 = d_2 a_2 - m_2 = 33 * 4 - 66 = 66`

`d_3 = (n - m_3^2) / d_2 = (6798 - 66^2) / 33 = 74`

`a_3 = floor((a_0 + m_3) / d_3) = floor((82 + 66) / 74) = 2`

`m_4 = d_3 * a_3 - m_3 = 74 * 2 - 66 = 82`

`d_4 = (n - m_4^2) / d_3 = (6798 - 6724) / 74 = 1`

`a_4 = floor((a_0 + m_4) / d_4) = floor((82+82)/1) = 164 = 2 * a_0`

So `sqrt(6798) = [82;2,4,2,164,2,4,2,164,...]`

`[82;2,4,2] = 82+1/(2+1/(4+1/2))`

`=82+1/(2+1/(9/2))`

`=82+1/(2+2/9)`

`=82+1/(20/9)`

`=82+9/20`

`=1649/20`

So `p = 1649` and `q = 20`

`p^2 = 2719201`

`n q^2 = 6798 * 400 = 2719200`

So `1649/20` will be a good first approximation for a Newton Raphson type method to find `sqrt(6798)` accurately with the minimum waste digits.