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

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(Solution)
 
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==Solution==
 
==Solution==
We look at x and y (mod3), since 21 is a multiple of 3.
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We look at <math>x</math> and <math>y \pmod{3}</math>, since <math>21</math> is a multiple of <math>3</math>.
  
* Case 1: x=0mod3
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* Case 1: <math>x\equiv 0\pmod{3}</math>
** Case 1a: y=0mod3: Then <math>x^2+4xy+y^2</math> is divisible by <math>3^2=9</math>, but 21 isn't.
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** Case 1a: <math>y\equiv 0\pmod{3}</math>: Then <math>x^2+4xy+y^2</math> is divisible by <math>3^2=9</math>, but <math>21</math> isn't.
** Case 1b: y=1mod3: Then the LHS is 1mod3, while the RHS isn't.
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** Case 1b: <math>y\equiv 1\pmod{3}</math>: Then the LHS is <math>1\pmod{3}</math>, while the RHS isn't.
** Case 1c: y=2mod3: Then the LHS is 1mod3, while the RHS isn't.
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** Case 1c: <math>y\equiv 2\pmod{3}</math>: Then the LHS is <math>1\pmod{3}</math>, while the RHS isn't.
 
* Case 2: x=1mod3
 
* Case 2: x=1mod3
 
** Case 2a: y=0mod3: This is equivalent to case 1b.
 
** Case 2a: y=0mod3: This is equivalent to case 1b.

Latest revision as of 23:38, 27 October 2015

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 $y \pmod{3}$, since $21$ is a multiple of $3$.

  • Case 1: $x\equiv 0\pmod{3}$
    • Case 1a: $y\equiv 0\pmod{3}$: Then $x^2+4xy+y^2$ is divisible by $3^2=9$, but $21$ isn't.
    • Case 1b: $y\equiv 1\pmod{3}$: Then the LHS is $1\pmod{3}$, while the RHS isn't.
    • Case 1c: $y\equiv 2\pmod{3}$: Then the LHS is $1\pmod{3}$, 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