Let $x$ and $y$ be integers with $x + y \ne 0$. Find all pairs $(x, y)$ such that $$\frac{x^2 + y^2}{x + y}= 10.$$

(Walther Janous)
1. Start with the given equation:
   \[
   \frac{x^2 + y^2}{x + y} = 10
   \]
   Multiply both sides by \(x + y\) to clear the fraction:
   \[
   x^2 + y^2 = 10(x + y)
   \]
2. Rearrange the equation:
   \[
   x^2 + y^2 - 10x - 10y = 0
   \]
3. Complete the square for both \(x\) and \(y\):
   \[
   x^2 - 10x + y^2 - 10y = 0
   \]
   \[
   (x^2 - 10x + 25) + (y^2 - 10y + 25) = 50
   \]
   \[
   (x - 5)^2 + (y - 5)^2 = 50
   \]
4. Recognize that \((x - 5)^2 + (y - 5)^2 = 50\) represents a circle centered at \((5, 5)\) with radius \(\sqrt{50} = 5\sqrt{2}\).

5. Identify the integer solutions to \((x - 5)^2 + (y - 5)^2 = 50\). The possible pairs \((a, b)\) such that \(a^2 + b^2 = 50\) are:
   \[
   (7, 1), (1, 7), (5, 5)
   \]
   and their negatives:
   \[
   (-7, -1), (-1, -7), (-5, -5)
   \]

6. Translate these pairs back to the original coordinates \((x, y)\):
   - For \((x - 5)^2 = 7^2\) and \((y - 5)^2 = 1^2\):
     \[
     x - 5 = \pm 7 \quad \text{and} \quad y - 5 = \pm 1
     \]
     This gives the pairs:
     \[
     (x, y) = (12, 6), (-2, 6), (12, 4), (-2, 4)
     \]
   - For \((x - 5)^2 = 5^2\) and \((y - 5)^2 = 5^2\):
     \[
     x - 5 = \pm 5 \quad \text{and} \quad y - 5 = \pm 5
     \]
     This gives the pairs:
     \[
     (x, y) = (10, 10), (0, 10), (10, 0)
     \]
     Note that \((x, y) = (0, 0)\) is not valid because \(x + y \neq 0\).

7. Therefore, the valid pairs \((x, y)\) are:
   \[
   (12, 6), (-2, 6), (12, 4), (-2, 4), (10, 10), (0, 10), (10, 0)
   \]

The final answer is \( \boxed{ (12, 6), (-2, 6), (12, 4), (-2, 4), (10, 10), (0, 10), (10, 0) } \).