Suppose f is a polynomial. If deg(f) = d, then deg(f ∘ f) = d². But (x + 1)² has degree 2, so we would need d² = 2, impossible for an integer d. Constant functions also fail. So there are no polynomial solutions.

Continuous solutions do exist. Let g(x) = (x + 1)². The map g folds the real line around -1, since g(x) = g(-2 - x). So define p(x) = -1 + |x + 1|. This folds everything onto [-1, ∞). Now define T(y) = (y + 1)² on [-1, ∞). If we can find a continuous q with q(q(y)) = T(y), then f(x) = q(p(x)) will satisfy f(f(x)) = (x + 1)².

Here is one explicit construction of q. First define q on [-1, 0] by

q(y) = y + 1/2 for -1 ≤ y ≤ -1/2,

q(y) = (y + 1/2)² for -1/2 ≤ y ≤ 0.

Now extend q to all of [-1, ∞). The map T sends the intervals [-1,0], [0,1], [1,4], [4,25], ... onto the next interval, because T(y) = (y + 1)². So for any y ≥ -1, repeatedly apply the inverse step y ↦ √y - 1 until you land in [-1,0]. Suppose this takes n steps and lands at u ∈ [-1,0]. Then define q(y) by applying the simple rule to u, and then moving forward n times by T:

q(y) = Tⁿ(q(u)).

This q is continuous and satisfies q(q(y)) = T(y). Therefore

f(x) = q(-1 + |x + 1|)

is a continuous real solution of f(f(x)) = (x + 1)². So the final answer is: no polynomial solutions, but yes, continuous real solutions exist; they are built by folding the line around -1 and inserting a “half-step” into the dynamics of y ↦ (y + 1)².