Difference between revisions of "2005 Alabama ARML TST Problems/Problem 8"
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==Solution== | ==Solution== | ||
+ | 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: <math>x\equiv 0\pmod{3}</math> |
+ | ** 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: <math>y\equiv 1\pmod{3}</math>: Then the LHS is <math>1\pmod{3}</math>, while the RHS isn't. | ||
+ | ** 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 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>: | ||
+ | <math>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</math> | ||
+ | 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 <math>x=3x_1+2</math> and <math>y=3y_1+2</math>: | ||
+ | |||
+ | <math>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</math> | ||
+ | |||
+ | But 21 isn't 6mod9, it's 3mod9. | ||
+ | |||
+ | Therefore, there are absolutely no solutions to the above equation. | ||
==See Also== | ==See Also== | ||
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− | [[Category: | + | [[Category:Intermediate Number Theory Problems]] |
Latest revision as of 23:38, 27 October 2015
Problem
Find the number of ordered pairs of integers which satisfy
Solution
We look at and , since is a multiple of .
- Case 1:
- Case 1a: : Then is divisible by , but isn't.
- Case 1b: : Then the LHS is , while the RHS isn't.
- Case 1c: : Then the LHS is , 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 and :
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 and :
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 |