Question

X^2+xy+y^2=37 and 3xy+2y^2=68

Answer 1

Given :   x² + xy + y² = 37  , 3xy + 2y² = 68

To find : Integral Solution

Solution:

x² + xy + y² = 37

3xy + 2y² = 68

x² + xy + y² = 37

=> (x - y)²  + 3xy = 37

=> 3xy  ≤  37

xy   ≤  12

(x - y)² = 0  =>  3xy  = 37 ( no integral solutions)

(x - y)² = 1  => 3xy = 36  => xy  = 12

Possible solution    

-3 , - 4   or  3 , 4  

3xy + 2y² = 68  is also satisfied

(x - y)² = 4  => 3xy = 33  => xy  = 11  

1 , 11 does not satisfy  (x - y)² = 4

no possible solution

(x - y)² = 9  => 3xy = 28    ( no integral solutions)=> xy  = 11

(x - y)² = 16  => 3xy = 21   => xy = 7

1 , 7 does not satisfy  (x - y)² = 16

no possible solution

(x - y)² = 25  => 3xy = 12    => xy = 4

1, 4  or   2 ,2 Does not satisfy  (x - y)² = 25

(x - y)² = 36  => 3xy = 1     ( no integral solutions)

Hence only integral solution

(3 , 4)   or (-3 , -4 )

3² + (3)(4) + 4²  = 37

3(3)(4) + 2(4)²  = 68

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