Let $f(x)=x^2-2x+1.$ For some constant $k, f(x+k) = x^2+2x+1$ for all real numbers $x.$ Determine the value of $k.$
1. Given the function \( f(x) = x^2 - 2x + 1 \), we can rewrite it in a factored form:
   \[
   f(x) = (x-1)^2
   \]

2. We are given that for some constant \( k \), the function \( f(x+k) \) is equal to \( x^2 + 2x + 1 \) for all real numbers \( x \). Therefore, we need to find \( k \) such that:
   \[
   f(x+k) = x^2 + 2x + 1
   \]

3. Substitute \( x+k \) into the function \( f \):
   \[
   f(x+k) = (x+k-1)^2
   \]

4. Expand the expression:
   \[
   (x+k-1)^2 = (x+k-1)(x+k-1) = x^2 + 2x(k-1) + (k-1)^2
   \]

5. We need this to be equal to \( x^2 + 2x + 1 \):
   \[
   x^2 + 2x(k-1) + (k-1)^2 = x^2 + 2x + 1
   \]

6. By comparing the coefficients of \( x \) and the constant terms on both sides of the equation, we get:
   \[
   2x(k-1) = 2x \quad \text{and} \quad (k-1)^2 = 1
   \]

7. From the coefficient of \( x \):
   \[
   2(k-1) = 2 \implies k-1 = 1 \implies k = 2
   \]

8. From the constant term:
   \[
   (k-1)^2 = 1 \implies k-1 = \pm 1
   \]
   This gives us two possible solutions:
   \[
   k-1 = 1 \implies k = 2
   \]
   \[
   k-1 = -1 \implies k = 0
   \]

9. However, since \( k = 0 \) does not satisfy the coefficient comparison \( 2(k-1) = 2 \), the only valid solution is:
   \[
   k = 2
   \]

The final answer is \( \boxed{2} \)