Solve the equation
$(x^2 + 2x + 1)^2+(x^2 + 3x + 2)^2+(x^2 + 4x +3)^2+...+(x^2 + 1996x + 1995)^2= 0$
1. Consider the given equation:
   \[
   (x^2 + 2x + 1)^2 + (x^2 + 3x + 2)^2 + (x^2 + 4x + 3)^2 + \cdots + (x^2 + 1996x + 1995)^2 = 0
   \]

2. Notice that each term in the sum is a square of a quadratic polynomial. Let's rewrite each term:
   \[
   (x^2 + kx + (k-1))^2 \quad \text{for} \quad k = 2, 3, \ldots, 1996
   \]

3. Observe that each quadratic polynomial can be factored:
   \[
   x^2 + kx + (k-1) = (x + 1)^2 + (k-1)
   \]

4. Therefore, the equation becomes:
   \[
   \sum_{k=2}^{1996} \left( (x + 1)^2 + (k-1) \right)^2 = 0
   \]

5. Factor out \((x + 1)^2\) from each term:
   \[
   \sum_{k=2}^{1996} \left( (x + 1)^2 + (k-1) \right)^2 = (x + 1)^2 \sum_{k=2}^{1996} \left( 1 + \frac{k-1}{(x + 1)^2} \right)^2
   \]

6. Simplify the inner sum:
   \[
   (x + 1)^2 \sum_{k=2}^{1996} \left( 1 + \frac{k-1}{(x + 1)^2} \right)^2
   \]

7. Notice that \(\left( 1 + \frac{k-1}{(x + 1)^2} \right)^2\) is always positive for any real \(x\) and \(k\). Therefore, the sum \(\sum_{k=2}^{1996} \left( 1 + \frac{k-1}{(x + 1)^2} \right)^2\) is positive for any real \(x\).

8. For the entire expression to be zero, \((x + 1)^2\) must be zero because the sum of positive terms cannot be zero unless each term is zero, which is not possible here.

9. Therefore, we must have:
   \[
   (x + 1)^2 = 0
   \]

10. Solving for \(x\):
    \[
    x + 1 = 0 \implies x = -1
    \]

The final answer is \(\boxed{-1}\)