# 2005 Alabama ARML TST Problems/Problem 8

## Problem

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

## 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.