Project Euler – Problem 66 Solution

Problem

Consider quadratic Diophantine equations of the form:

x2 – Dy2 = 1

For example, when D=13, the minimal solution in x is 6492 – 13×1802 = 1.

It can be assumed that there are no solutions in positive integers when D is square.

By finding minimal solutions in x for D = {2, 3, 5, 6, 7}, we obtain the following:

32 – 2×22 = 1

22 – 3×12 = 1

92 – 5×42 = 1

52 – 6×22 = 1

82 – 7×32 = 1

Hence, by considering minimal solutions in x for D <= 7, the largest x is obtained when D=5.

Find the value of D <= 1000 in minimal solutions of x for which the largest value of x is obtained.

Solution

// define function to get the continued fractions, a0,a1,a2,...an
let continuedFraction D =
    // define functions for working out the nth term of P, Q and a sequence
    let getPn an' qn' pn' = an' * qn' - pn'
    let getQn D pn qn' = (D - pown pn 2) / qn'
    let getAn a0 pn qn = (a0 + pn) / qn

    // work out the initial terms
    let a0 = bigint(sqrt(double(D)))
    let p1, q1 = a0, D-a0*a0
    let a1 = getAn a0 p1 q1
    let initial = (p1, q1, a1)

    Seq.unfold (fun (pn', qn', an') ->
        let pn = getPn an' qn' pn'
        let qn = getQn D pn qn'
        let an = getAn a0 pn qn
        Some((pn', qn', an'), (pn, qn, an))) initial
    |> Seq.map (fun (pn, qn, an) -> an)
    |> Seq.append [a0]

// define function to get the continued fraction convergents
// e.g. for D = 7: 2/1, 3/1, 5/2, 8/3, ...
let continuedFractionConvergents D =
    let getN an n' n'' = an * n' + n''

    // work out the initial terms
    let fractions = continuedFraction D
    let a0 = Seq.head fractions
    let p0, p1 = a0, a0 * (Seq.nth 1 fractions) + 1I
    let q0, q1 = 1I, Seq.nth 1 fractions
    let initial = (p1, q1, p0, q0)

    Seq.scan (fun (pn', qn', pn'', qn'') an ->
        let pn = getN an pn' pn''
        let qn = getN an qn' qn''
        (pn, qn, pn', qn')) initial (fractions |> Seq.skip 2)
    |> Seq.map (fun (pn, qn, pn', qn') -> (pn, qn))
    |> Seq.append [(p0, q0)]

let answer =
    [1I..1000I]
    |> List.filter (fun d -> sqrt(double(d)) % 1.0 <> 0.0)
    |> List.maxBy (fun d ->
        continuedFractionConvergents d
        |> Seq.filter (fun (x, y) -> x*x - d*y*y = 1I)
        |> Seq.map fst
        |> Seq.head)

After spending much time reading the information on Pell’s equation on mathworld and wikipedia, it turns out that the best way to solve the equation for a given D is to work out the continued fraction convergents to the square root of D.

For example, for D = 7, the equation becomes:

image

The fundamental solution (the first pair of x, y which satisfies the equation) can be obtained by going through the sequence of convergents for the square root of 7:

image image
image image
image image
image image
image image

So how do we workout the values of the convergents? The continued fraction for a number is usually written as

image

The first ten values generated by the continued fraction for the square root of 2 are:

1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378, …

These values are called continued fraction convergents, getting closer to the true value of square root 2 with each step..

So, to find the convergents to the square root of D, we first need to find the values of a:

clip_image002[8]

clip_image004

clip_image002[10]

clip_image002[6]

Once we have the sequence of values for a, we can then find out the convergent values:

clip_image002[12]

clip_image004[6]

The rest is simple, for each value of D between 1 and 1000, go through the convergents to find the fundamental solution and return the max value of x in all the fundamental solutions.

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Yan Cui

I’m an AWS Serverless Hero and the author of Production-Ready Serverless. I have run production workload at scale in AWS for nearly 10 years and I have been an architect or principal engineer with a variety of industries ranging from banking, e-commerce, sports streaming to mobile gaming. I currently work as an independent consultant focused on AWS and serverless.

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