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Odd period square roots

Problem 64

Published on Friday, 27th February 2004, 06:00 pm; Solved by 10991

All square roots are periodic when written as continued fractions and can be written in the form:

√N = a0 +
1
  a1 +
1
    a2 +
1
      a3 + ...

For example, let us consider √23:

√23 = 4 + √23 — 4 = 4 + 
1
 = 4 + 
1
 
1
√23—4
  1 + 
√23 – 3
7

If we continue we would get the following expansion:

√23 = 4 +
1
  1 +
1
    3 +
1
      1 +
1
        8 + ...

The process can be summarised as follows:

a0 = 4,  
1
√23—4
 = 
√23+4
7
 = 1 + 
√23—3
7
a1 = 1,  
7
√23—3
 = 
7(√23+3)
14
 = 3 + 
√23—3
2
a2 = 3,  
2
√23—3
 = 
2(√23+3)
14
 = 1 + 
√23—4
7
a3 = 1,  
7
√23—4
 = 
7(√23+4)
7
 = 8 +  √23—4
a4 = 8,  
1
√23—4
 = 
√23+4
7
 = 1 + 
√23—3
7
a5 = 1,  
7
√23—3
 = 
7(√23+3)
14
 = 3 + 
√23—3
2
a6 = 3,  
2
√23—3
 = 
2(√23+3)
14
 = 1 + 
√23—4
7
a7 = 1,  
7
√23—4
 = 
7(√23+4)
7
 = 8 +  √23—4

It can be seen that the sequence is repeating. For conciseness, we use the notation √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:

√2=[1;(2)], period=1
√3=[1;(1,2)], period=2
√5=[2;(4)], period=1
√6=[2;(2,4)], period=2
√7=[2;(1,1,1,4)], period=4
√8=[2;(1,4)], period=2
√10=[3;(6)], period=1
√11=[3;(3,6)], period=2
√12= [3;(2,6)], period=2
√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?