Find all integers $ x$ such that $ x(x\plus{}1)(x\plus{}7)(x\plus{}8)$ is a perfect square
It's a nice problem ...hope you enjoy it!

Daniel
To find all integers \( x \) such that \( x(x + 1)(x + 7)(x + 8) \) is a perfect square, we start by setting:
\[ P(x) = x(x + 1)(x + 7)(x + 8) \]

1. **Rewrite the expression in terms of a new variable:**
   Let \( b = x^2 + 8x \). Then:
   \[ P(x) = (x^2 + 8x)(x^2 + 8x + 7) = b(b + 7) \]

2. **Set up the equation for a perfect square:**
   We need \( b(b + 7) = a^2 \) for some integer \( a \). This can be rewritten as:
   \[ 4b(b + 7) = 4a^2 \]
   \[ 4b^2 + 28b = 4a^2 \]
   \[ 4b^2 + 28b + 49 - 49 = 4a^2 \]
   \[ (2b + 7)^2 - 49 = 4a^2 \]
   \[ (2b + 7)^2 - 4a^2 = 49 \]

3. **Factor the difference of squares:**
   \[ (2b + 7 - 2a)(2b + 7 + 2a) = 49 \]

4. **Solve the factor pairs of 49:**
   The factor pairs of 49 are \( (1, 49), (7, 7), (-1, -49), (-7, -7) \). We solve for each pair:
   
   - For \( 2b + 7 - 2a = 1 \) and \( 2b + 7 + 2a = 49 \):
     \[ 4b + 14 = 50 \]
     \[ 4b = 36 \]
     \[ b = 9 \]
     \[ 2a = 24 \]
     \[ a = 12 \]
   
   - For \( 2b + 7 - 2a = 7 \) and \( 2b + 7 + 2a = 7 \):
     \[ 4b + 14 = 14 \]
     \[ 4b = 0 \]
     \[ b = 0 \]
     \[ 2a = 0 \]
     \[ a = 0 \]
   
   - For \( 2b + 7 - 2a = -1 \) and \( 2b + 7 + 2a = -49 \):
     \[ 4b + 14 = -50 \]
     \[ 4b = -64 \]
     \[ b = -16 \]
     \[ 2a = -24 \]
     \[ a = -12 \]
   
   - For \( 2b + 7 - 2a = -7 \) and \( 2b + 7 + 2a = -7 \):
     \[ 4b + 14 = -14 \]
     \[ 4b = -28 \]
     \[ b = -7 \]
     \[ 2a = -14 \]
     \[ a = -7 \]

5. **Find corresponding \( x \) values:**
   - For \( b = 9 \):
     \[ x^2 + 8x = 9 \]
     \[ x^2 + 8x - 9 = 0 \]
     \[ (x + 9)(x - 1) = 0 \]
     \[ x = -9 \text{ or } x = 1 \]
   
   - For \( b = 0 \):
     \[ x^2 + 8x = 0 \]
     \[ x(x + 8) = 0 \]
     \[ x = 0 \text{ or } x = -8 \]
   
   - For \( b = -16 \):
     \[ x^2 + 8x = -16 \]
     \[ x^2 + 8x + 16 = 0 \]
     \[ (x + 4)^2 = 0 \]
     \[ x = -4 \]
   
   - For \( b = -7 \):
     \[ x^2 + 8x = -7 \]
     \[ x^2 + 8x + 7 = 0 \]
     \[ (x + 7)(x + 1) = 0 \]
     \[ x = -7 \text{ or } x = -1 \]

Conclusion:
The integer solutions for \( x \) are:
\[ x = -9, -8, -7, -4, -1, 0, 1 \]

The final answer is \(\boxed{-9, -8, -7, -4, -1, 0, 1}\)