#64 - Odd Period Square Roots
All square roots are periodic when written as continued fractions and can be written in the form:
\[\sqrt{N}=a_0 + \dfrac 1 {a_1 + \dfrac 1 {a_2 + \dfrac 1 {a_3 + \dots}}}\]For example, let us consider \(\sqrt{23}:\)
\[\sqrt{23} = 4 + \sqrt{23}-4=4 + \dfrac 1 {\dfrac 1 {\sqrt{23}-4}} = 4+\dfrac 1 {1 + \dfrac{\sqrt{23}-3}7}\]If we continue we would get the following expansion:
\[\sqrt{23}=4 + \dfrac 1 {1 + \dfrac 1 {3+ \dfrac 1 {1 + \dfrac 1 {8+ \dots}}}}\]The process can be summarised as follows:
\[\begin{align} \quad \quad a_0 &= 4, \frac 1 {\sqrt{23}-4}=\frac {\sqrt{23}+4} 7=1+\frac {\sqrt{23}-3} 7 \\ \quad \quad a_1 &= 1, \frac 7 {\sqrt{23}-3}=\frac {7(\sqrt{23}+3)} {14}=3+\frac {\sqrt{23}-3} 2 \\ \quad \quad a_2 &= 3, \frac 2 {\sqrt{23}-3}=\frac {2(\sqrt{23}+3)} {14}=1+\frac {\sqrt{23}-4} 7 \\ \quad \quad a_3 &= 1, \frac 7 {\sqrt{23}-4}=\frac {7(\sqrt{23}+4)} 7=8+\sqrt{23}-4 \\ \quad \quad a_4 &= 8, \frac 1 {\sqrt{23}-4}=\frac {\sqrt{23}+4} 7=1+\frac {\sqrt{23}-3} 7 \\ \quad \quad a_5 &= 1, \frac 7 {\sqrt{23}-3}=\frac {7 (\sqrt{23}+3)} {14}=3+\frac {\sqrt{23}-3} 2 \\ \quad \quad a_6 &= 3, \frac 2 {\sqrt{23}-3}=\frac {2(\sqrt{23}+3)} {14}=1+\frac {\sqrt{23}-4} 7 \\ \quad \quad a_7 &= 1, \frac 7 {\sqrt{23}-4}=\frac {7(\sqrt{23}+4)} {7}=8+\sqrt{23}-4 \end{align}\]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:
\(\quad \quad \sqrt{2}=[1;(2)]\), period=\(1\)
\(\quad \quad \sqrt{3}=[1;(1,2)]\), period=\(2\)
\(\quad \quad \sqrt{5}=[2;(4)]\), period=\(1\)
\(\quad \quad \sqrt{6}=[2;(2,4)]\), period=\(2\)
\(\quad \quad \sqrt{7}=[2;(1,1,1,4)]\), period=\(4\)
\(\quad \quad \sqrt{8}=[2;(1,4)]\), period=\(2\)
\(\quad \quad \sqrt{10}=[3;(6)]\), period=\(1\)
\(\quad \quad \sqrt{11}=[3;(3,6)]\), period=\(2\)
\(\quad \quad \sqrt{12}=[3;(2,6)]\), period=\(2\)
\(\quad \quad \sqrt{13}=[3;(1,1,1,1,6)]\), 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?
Problem text © Project Euler, licensed under CC BY-NC-SA 4.0. Original: projecteuler.net/problem=64. Published Friday, 27th February 2004, 06:00 pm. Solved by 25,465 members at time of mirroring.
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Prerequisites
Lessons that prepare you:
1.3 Proof Techniques: Direct, Contrapositive, and Contradiction · 1.1 Sets, Functions, and Relations · 19.12 Continued Fractions, Pell Equations, and Diophantine Approximation · 19.1 Divisibility, GCD, and the Euclidean Algorithm · 19.4 Exact Arithmetic: Big Integers, Rationals, and Floating-Point Traps · 19.3 Modular Arithmetic, Inverses, and Fast Exponentiation · 19.6 Recurrence Relations and Generating Functions · 19.10 Search: Backtracking, Branch-and-Bound, Binary Search, Meet-in-the-Middle · 19.2 Primes, Sieves, and Integer Factorization · 2.1 Functions, Limits, and Continuity · 2.7 Sequences, Series, Convergence, and Power Series · 3.1 Vectors, Multivariable Functions, and Level Sets · 4.2 Linear Maps, Matrices, Rank, and the Null Space
Concepts: algebra geometry number-theory sequences-series
Likely techniques: continued-fractions exact-rational
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- What exactly is the input to problem 64? Is it a bound (10000), a supplied dataset, or a definition you must generate from?
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