Find the total number of solutions to the equation `x^2 + y^2 = 2015` where both x and y are integers.

Answer

To find the total number of integer solutions to the equation \( x^2 + y^2 = 2015 \), we can follow these steps:

### Step 1: Determine the form of the equation
The equation \( x^2 + y^2 = 2015 \) suggests that we are looking for pairs of integers \( (x, y) \) such that their squares sum to 2015.

### Step 2: Check the modulo condition
We can analyze the equation modulo 4. The possible values of \( x^2 \) and \( y^2 \) modulo 4 are:
- \( 0 \) if \( x \) or \( y \) is even,
- \( 1 \) if \( x \) or \( y \) is odd.

Thus, the possible sums \( x^2 + y^2 \) modulo 4 can be:
- \( 0 + 0 = 0 \)
- \( 0 + 1 = 1 \)
- \( 1 + 0 = 1 \)
- \( 1 + 1 = 2 \)

Now, we compute \( 2015 \mod 4 \):
\[ 2015 \div 4 = 503 \quad \text{(remainder 3)} \]
So, \( 2015 \equiv 3 \mod 4 \).

### Step 3: Analyze the implications
Since \( x^2 + y^2 \) can only yield results of \( 0, 1, \) or \( 2 \) modulo 4, it is impossible for \( x^2 + y^2 \) to be congruent to \( 3 \mod 4 \). Therefore, there cannot be any integer solutions to the equation \( x^2 + y^2 = 2015 \).

### Conclusion
Thus, the total number of integer solutions to the equation \( x^2 + y^2 = 2015 \) is **0**.

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