Transcript Document

Isosceles Right Angled Triangle
The three sides cannot be all integers
1
2
1
Find a right angled triangle with integral
sides near to be isosceles
such as
5
4
3
Question
• The difference between two legs is 1
• What is the next triangle?
• Are there sequences of the legs with ratio tending to
1:1?
The three sides should be
a, a + 1 and a + k
where a, k are positive integers
a+1
and k > 1
a
Relation between a and k
• a2 + (a + 1)2 = (a + k)2
• a2 – 2(k – 1)a – (k2 – 1) = 0
• a  k  1  2k (k  1)
•
a  k  1  2k (k  1) as a > 0
Discriminant
• 2k(k – 1) is a perfect square
• For k, k – 1, one is odd, one is even
• Since k, k – 1 are relatively prime,
odd one = x2, even one = 2y2
• Then x2 – 2y2 = 1
• 2k(k – 1) = 4x2y2
Find the three sides
a  k  1  2k (k  1)
 k 1  4x y
2
 k  1 2 xy
2
Find the three sides
If k is odd, then k = x2, then the three sides are:
2y2 + 2xy, x2 + 2xy, x2 + 2xy + 2y2
If k is even, then k = 2y2, then the three sides are:
x2 + 2xy, 2y2 + 2xy, x2 + 2xy + 2y2
Both cases have the same form.
Pell’s Equation
• x2 – 2y2 = 1
• How to solve it?
• Making use the expansion of 2
into infinite continued fraction
Continued Fraction
2  1  2 1
2 1
 1
2 1
1
 1
2  2 1
1
 1
2
1
1
2
2  ...
Sequence of the Fractions
f1  1,
1 3
f2  1  ,
2 2
1
7
f3  1 

1 5
2
2
Sequence of the Fractions
{fn} is generated by the sequence {1, 2, 2, 2, …. }
hn
fn 
kn
where h1 = 1, h2 = 3,
k1 = 1, k2 = 2
hn = hn-2 + 2hn-1,
kn = kn-2 + 2kn-1 for n > 2
Solution of Pell’s Equation x2 – 2y2 = 1
• (hn, kn) is a solution of x2 – 2y2 = 1 when n is even.
• (hn, kn) is a solution of x2 – 2y2 = -1 when n is odd.
• This is the full solution set.
Sequences of the Sides
n
1
2
3
4
5
x = hn
1
3
7
17
41
y = kn
1
2
5
12
29
an=x2+2xy
3
21
119
697
4059
bn=2y(x+y)
4
20
120
696
4060
x2+2xy+2y2
5
29
169
985
5741
Properties of Sequences
hn
lim
 2
n  k
n
an
lim
1
n b
n
The triangle tends to be isosceles