How many integer pairs (x,y) satisfy `x^(2) + 4y^(2) -2xy -2x - 4y -8=0` ?

Answer

To solve the equation \( x^2 + 4y^2 - 2xy - 2x - 4y - 8 = 0 \) for integer pairs \((x, y)\), we can follow these steps:

### Step 1: Rearranging the Equation
We start with the given equation:
\[ x^2 + 4y^2 - 2xy - 2x - 4y - 8 = 0 \]
We can rearrange it to isolate the quadratic terms:
\[ x^2 - 2xy + 4y^2 - 2x - 4y - 8 = 0 \]

### Step 2: Completing the Square
Next, we can try to complete the square for the \(x\) and \(y\) terms. We rewrite the equation:
\[ (x^2 - 2xy + 4y^2) - 2x - 4y - 8 = 0 \]
Now, we can group the \(x\) terms:
\[ (x - 2y)^2 + 3y^2 - 2x - 4y - 8 = 0 \]

### Step 3: Substituting Variables
Let \(a = x - 2y\) and \(b = y\). Then, we can rewrite the equation:
\[ a^2 + 3b^2 - 2(2b + 4) = 0 \]
This simplifies to:
\[ a^2 + 3b^2 = 48 \]

### Step 4: Finding Integer Solutions
Now we need to find integer pairs \((a, b)\) such that:
\[ a^2 + 3b^2 = 48 \]
We can check for integer values of \(b\):
- If \(b = 0\), \(a^2 = 48\) (not an integer).
- If \(b = 1\), \(a^2 + 3 = 48 \Rightarrow a^2 = 45\) (not an integer).
- If \(b = 2\), \(a^2 + 12 = 48 \Rightarrow a^2 = 36 \Rightarrow a = \pm 6\).
- If \(b = 3\), \(a^2 + 27 = 48 \Rightarrow a^2 = 21\) (not an integer).
- If \(b = 4\), \(a^2 + 48 = 48 \Rightarrow a^2 = 0 \Rightarrow a = 0\).

### Step 5: Listing the Solutions
From the above calculations, we have the following pairs:
1. \( (6, 2) \)
2. \( (-6, 2) \)
3. \( (0, 4) \)

Now we can convert back to \((x, y)\):
- For \(b = 2\) (which is \(y\)):
- \(x = a + 2b = 6 + 4 = 10\) and \(y = 2\) gives \((10, 2)\)
- \(x = -6 + 4 = -2\) and \(y = 2\) gives \((-2, 2)\)
- For \(b = 4\):
- \(x = 0 + 8 = 8\) and \(y = 4\) gives \((8, 4)\)

### Final Count of Integer Pairs
The integer pairs that satisfy the equation are:
1. \( (10, 2) \)
2. \( (-2, 2) \)
3. \( (8, 4) \)

Thus, the total number of integer pairs \((x, y)\) that satisfy the equation is **3**.