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Integral solution of the biquadratic Equation<br />

II. METHOD OF ANALYSIS<br />

The non-homogeneous biquadratic diophantine equation with four unknowns is<br />

4 4<br />

2<br />

2 2<br />

x y ( k 1)( z w )<br />

(1)<br />

It is worth to note that (1) is satisfied by the following non-zero distinct integer quadruples<br />

2 2<br />

2 2<br />

2 2 2<br />

2 2 2<br />

(( a b )( 1 k ), ( a b )( 1 k ), ( a b ) ( 2 k 1), ( a b ) ( 2 k 1)) and<br />

2<br />

2<br />

4<br />

4<br />

(10 b ( k 1), 10 b ( k 1), 100 b ( 2 k 1), 100 b ( 2 k 1)) .<br />

However we have some more patterns of solutions to (1) which are illustrated below<br />

To start with,<br />

the substitution of the transformations<br />

2<br />

2<br />

x u v , y u v , z 2 uv , w 2 uv <br />

(2)<br />

2 2<br />

2<br />

2<br />

in (1), leads to u v ( k 1) <br />

(3)<br />

2.1 Pattern 1<br />

2 2<br />

Let a b<br />

(4)<br />

Substituting (4) in (3) and using the method of factorization, define<br />

2<br />

iv ( k i )( a ib )<br />

(5)<br />

u<br />

Equating real and imaginary parts in (5), we have<br />

2<br />

2<br />

u k ( a b ) 2 ab<br />

(6)<br />

2 2<br />

v ( a b ) 2 abk<br />

Substituting (4) , (6) in (2) and simplifying, the corresponding values of x,y,z and w are represented by<br />

x ( k , a , b )<br />

2 2<br />

2 2<br />

k (a b 2 ab ) (a b 2 ab )<br />

y ( k , a , b )<br />

<br />

2<br />

k (a<br />

2<br />

b<br />

2<br />

2 ab ) (a<br />

2<br />

b<br />

2 ab )<br />

z ( k , a , b )<br />

<br />

2 2<br />

4 k ab (a<br />

2<br />

4<br />

b ) 2 k (a<br />

4<br />

b<br />

2 2<br />

2<br />

6 a b ) 4 ab (a<br />

2 2<br />

b ) (a<br />

2 2<br />

b )<br />

Properties<br />

w ( k , a , b )<br />

2 2 2<br />

4 4 2 2<br />

2 2 2 2 2<br />

4 k ab (a b ) 2 k (a b 6 a b ) 4 ab (a b ) (a b )<br />

4<br />

1. 4 z ( k , a 1, a ) w ( k , a 1, a ) 192 PT 96 T 24 PR <br />

is a cubical integer.<br />

2.Each of the following is a nasty number:<br />

(i) 3(z(k,a,b)-w(k,a,b))<br />

(ii) x(k,ka,a)<br />

(iii) -6(x(k,a,a)+y(k,a,a)).<br />

a<br />

2<br />

3<br />

5<br />

3. z k , a , b ) w ( k , a , b ) 48 ( k 1) P k (32 P 48 F 8t<br />

4) 0(mod 28 )<br />

( a 1<br />

a<br />

4 , a , 4 9 , a<br />

n<br />

n<br />

4. x k ,2 ,1) y ( k ,2 ,1) 12 kJ KY j o (mod )<br />

( k<br />

n<br />

2 n 2 n<br />

3<br />

2<br />

5. x k ,1, b ) y ( k ,1, b ) 48 kP ( k 1)( 24 PT 9OH<br />

t 18 t 36 t 1)<br />

( <br />

b 1 b<br />

b 19 , b 6 , b<br />

4 , b<br />

2.2 Pattern2:<br />

(3) is written as<br />

Write „1‟ as<br />

a<br />

2 2<br />

2<br />

2<br />

u v ( k 1) * 1<br />

(7)<br />

( 3 4 i )( 3 4 i )<br />

1=<br />

25<br />

Using (4) and (8) in (7) and employing the method of factorization, define<br />

(3 4 i )<br />

u<br />

5<br />

Equating real and imaginary parts of (9) we get<br />

2<br />

iv ( k i ) ( a ib )<br />

(9)<br />

a<br />

(8)<br />

www.<strong>ijcer</strong>online.com ||May ||2013|| Page 29

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