2005 Alabama ARML TST Problems/Problem 8

Problem

Find the number of ordered pairs of integers $(x,y)$ which satisfy $x^2+4xy+y^2=21$ .

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 $y=3y_1+1$ :

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 $y=3y_1+2$ :

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

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