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Difference between revisions of "2005 Alabama ARML TST Problems/Problem 8"

(Could someone fix this up? Because it looks messy.)
m (grammar and repetition fixes.)
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** Case 1c: y=2mod3: Then the LHS is 1mod3, while the RHS isn't.
 
** Case 1c: y=2mod3: Then the LHS is 1mod3, while the RHS isn't.
 
* Case 2: x=1mod3
 
* Case 2: x=1mod3
−
** Case 2a: y=0mod3: Then the LHS is 1mod3 while the RHS isn't.
+
** Case 2a: y=0mod3: This is equivalent to case 1b.
 
** Case 2b: y=1mod3: We let <math>x=3x_1+1</math> and <math>y=3y_1+1</math>:
 
** Case 2b: y=1mod3: We let <math>x=3x_1+1</math> and <math>y=3y_1+1</math>:
  
Line 19: Line 19:
 
** Case 2c: y=2mod3: Then the LHS is 1mod3 while the RHS isn't.
 
** Case 2c: y=2mod3: Then the LHS is 1mod3 while the RHS isn't.
 
* Case 3: x=2mod3
 
* Case 3: x=2mod3
−
** Case 3a: y=0mod3: This is equivalent to case 1c
+
** Case 3a: y=0mod3: This is equivalent to case 1c.
−
** Case 3b: y=1mod3: Equivalent to case 2c
+
** Case 3b: y=1mod3: This is equivalent to case 2c.
 
** Case 3c: y=2mod3: We let <math>x=3x_1+2</math> and <math>y=3y_1+2</math>:
 
** Case 3c: y=2mod3: We let <math>x=3x_1+2</math> and <math>y=3y_1+2</math>:
  

Revision as of 10:53, 18 June 2008

Problem

Find the number of ordered pairs of integers $(x,y)$ which satisfy

$x^2+4xy+y^2=21$.

Solution

We look at x and y (mod3), since 21 is a multiple of 3.

  • Case 1: x=0mod3
    • Case 1a: y=0mod3: Then $x^2+4xy+y^2$ is divisible by $3^2=9$, but 21 isn't.
    • Case 1b: y=1mod3: Then the LHS is 1mod3, while the RHS isn't.
    • Case 1c: y=2mod3: Then the LHS is 1mod3, while the RHS isn't.
  • Case 2: x=1mod3
    • Case 2a: y=0mod3: This is equivalent to case 1b.
    • Case 2b: y=1mod3: We let $x=3x_1+1$ and $y=3y_1+1$:

$x^2+4xy+y^2=21=(3x_1+1)^2+4(3x_1+1)(3y_1+1)+(3y_1+1)^2=9(x^2+y^2+4x_1y_1+2x_1+2y_1)+6$

But 21 isn't 6mod9, it's 3mod9.

    • Case 2c: y=2mod3: Then the LHS is 1mod3 while the RHS isn't.
  • Case 3: x=2mod3
    • Case 3a: y=0mod3: This is equivalent to case 1c.
    • Case 3b: y=1mod3: This is equivalent to case 2c.
    • Case 3c: y=2mod3: We let $x=3x_1+2$ and $y=3y_1+2$:

$x^2+4xy+y^2=21=(3x_1+2)^2+4(3x_1+2)(3y_1+2)+(3y_1+2)^2=9(x_1^2+y_1^2+4x_1y_1+4x_1+4y_1+2)+6$

But 21 isn't 6mod9, it's 3mod9.

Therefore, there are absolutely no solutions to the above equation.

See Also

2005 Alabama ARML TST (Problems)
Preceded by:
Problem 7 		Followed by:
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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