1. Show that k=15 is congruent integer.
where 0.5xy=km^2,
x=r(u^2 - v^2)
y=r(2uv)
z=r(u^2 + v^2)
x^2 + y^2 = z^2
2. how i solve Pell's equation using bhaskara II then brahmagupta method for x^2 - 67y^2 =1
Please help, exam on monday... :)
#2
I don't know brahmagupta's method, but I remembered a method that uses continued fractions.
It is a rather easy method compared to some of the other methods that are out there
It involves finding the continued fraction terms of ?67
which are
8; 5,2,1,1,7,1,1,2,5,16
Then I complete a complicated chart and found
x = 48842 , y = 5967 to be a solution
(48842)^2 - 67(5967)^2 = 1
continued fractions are an amazing topic
and if you want to learn more here is start:
http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/cfINTRO.html
an remarkable page!!!
Yea i know... but my lecturer wants us to solve it using that method.... actually i know how to use brahmagupta method... but i just cant get bhaskara II to transform into something like brahmagupta
1. To show that k=15 is a congruent integer, we need to find values for the variables that satisfy the given equations and show that the resulting k is an integer. Let's break down the steps:
Step 1: Start with the equation 0.5xy = km^2. Set k=15 and solve the equation for x, y, and m.
0.5xy = 15m^2
Multiply both sides by 2 to eliminate the fraction:
xy = 30m^2
Step 2: Next, substitute x, y, and z from the equations x = r(u^2 - v^2), y = r(2uv), and z = r(u^2 + v^2) into the equation x^2 + y^2 = z^2.
(r(u^2 - v^2))^2 + (r(2uv))^2 = (r(u^2 + v^2))^2
Expand and simplify:
r^2(u^4 - 2u^2v^2 + v^4) + 4r^2(u^2v^2) = r^2(u^4 + 2u^2v^2 + v^4)
Cancel out the r^2 and simplify:
u^4 - 2u^2v^2 + v^4 + 4u^2v^2 = u^4 + 2u^2v^2 + v^4
Simplify further:
2u^2v^2 = 6u^2v^2
Step 3: Now, solve for u and v:
2u^2v^2 - 6u^2v^2 = 0
-4u^2v^2 = 0
Since -4u^2v^2 = 0, it means that either u or v must be zero.
Step 4: Assume u = 0:
2(0)^2v^2 - 6(0)^2v^2 = 0
0 - 0 = 0
This equation holds true when u = 0.
Step 5: Assume v = 0:
2u^2(0)^2 - 6u^2(0)^2 = 0
0 - 0 = 0
This equation also holds true when v = 0.
Therefore, either u = 0 or v = 0, and the equation 0.5xy = 15m^2 holds true. Thus, k=15 is a congruent integer.
2. To solve Pell's equation x^2 - 67y^2 = 1 using Bhaskara II and Brahmagupta method:
Step 1: Set up the equation as x^2 - 67y^2 = 1.
Step 2: Use Bhaskara II's method for solving Pell's equation, which involves finding the continued fraction representation of the square root of the non-square term (in this case, 67).
Calculate the square root of 67: √67 ≈ 8.1853
The whole part is 8, and the decimal part is 0.1853.
Now, represent this decimal part as a continued fraction: 1 / 0.1853 = 5.3981 ≈ 5 + 1 / 2.602
Continuing this process, we get the continued fraction representation: 8.1853 ≈ [8; 2, 8, 2, 8, ...]
Step 3: Now, using the Brahmagupta method, the basic form of the solution can be found by taking the convergents of the continued fraction. The continued fraction representation [8; 2, 8, 2, 8, ...] yields the convergents:
Convergent 1: 8/1 = 8
Convergent 2: 17/2 ≈ 8.5
Convergent 3: 153/18 ≈ 8.5
Convergent 4: 1216/143 ≈ 8.5
Convergent 5: 10345/1211 ≈ 8.5
Step 4: From the convergents, select the one that satisfies x^2 - 67y^2 = 1. In this case, 9^2 - 67(2^2) = 1, so the solution is x = 9 and y = 2.
Therefore, x = 9 and y = 2 is a solution to Pell's equation x^2 - 67y^2 = 1.