# Problem 64: Odd period square roots

![graphic](img064.gif)

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 – 37 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+47  = 1
+  √23—37 a1 = 1,   7√23—3  =  7(√23+3)14  = 3 +  √23—32 a2 = 3,  
2√23—3  =  2(√23+3)14  = 1 +  √23—47 a3 = 1,   7√23—4  =  7(√23+4)7  = 8
+  √23—4 a4 = 8,   1√23—4  =  √23+47  = 1 +  √23—37 a5 = 1,   7√23—3  = 
7(√23+3)14  = 3 +  √23—32 a6 = 3,   2√23—3  =  2(√23+3)14  = 1 +  √23—47
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?