Find all integers n greater than or equal to $3$ such that $\sqrt{\frac{n^2 - 5}{n + 1}}$ is a rational number.
To find all integers \( n \geq 3 \) such that \( \sqrt{\frac{n^2 - 5}{n + 1}} \) is a rational number, we need to ensure that the expression under the square root is a perfect square. Let's denote this rational number by \( k \), so we have:

\[
\sqrt{\frac{n^2 - 5}{n + 1}} = k
\]

Squaring both sides, we get:

\[
\frac{n^2 - 5}{n + 1} = k^2
\]

Multiplying both sides by \( n + 1 \), we obtain:

\[
n^2 - 5 = k^2 (n + 1)
\]

Rearranging the equation, we get:

\[
n^2 - k^2 n - k^2 - 5 = 0
\]

This is a quadratic equation in \( n \). For \( n \) to be an integer, the discriminant of this quadratic equation must be a perfect square. The discriminant \( \Delta \) of the quadratic equation \( n^2 - k^2 n - (k^2 + 5) = 0 \) is given by:

\[
\Delta = (k^2)^2 - 4 \cdot 1 \cdot (-k^2 - 5) = k^4 + 4k^2 + 20
\]

For \( n \) to be an integer, \( \Delta \) must be a perfect square. Let's denote \( \Delta \) by \( m^2 \):

\[
k^4 + 4k^2 + 20 = m^2
\]

We need to solve this Diophantine equation for integer values of \( k \) and \( m \). However, we can also approach the problem by considering the possible values of \( \gcd(n^2 - 5, n + 1) \).

We observe that:

\[
\gcd(n^2 - 5, n + 1) = \gcd(n^2 - 5 - n(n + 1), n + 1) = \gcd(-n - 5, n + 1) = \gcd(-4, n + 1)
\]

So, the possible values of \( \gcd(n^2 - 5, n + 1) \) are 1, 2, or 4. Given that we want the whole root to be rational, we need the numerator and the denominator to be squares. Thus, we consider the following cases:

### Case 1: \( a = 1 \)
\[
n^2 - 5 = s^2 \quad \text{and} \quad n + 1 = t^2
\]

Evaluating modulo 4, we see that \( n \) must be congruent to 1 or -1 by the first equation and 0 or -1 by the second equation. Therefore, \( n \) must be -1. However, this does not satisfy the condition \( n \geq 3 \).

### Case 2: \( a = 2 \)
\[
n^2 - 5 = 2s^2 \quad \text{and} \quad n + 1 = 2t^2
\]

Evaluating modulo 5, we get that \( n \) is -1, 1, or 2 by the second equation, and 0 by the first equation. There are no common elements, so we get a contradiction.

### Case 3: \( a = 4 \)
\[
n^2 - 5 = 4s^2 \quad \text{and} \quad n + 1 = 4t^2
\]

Observe that in the first equation, we have two perfect squares separated by 5, so they must be 4 and 9. Thus, \( n = 3 \) and \( s = 1 \). Then, the second equation gives \( t = 1 \).

Thus, the only solution is \( n = 3 \).

The final answer is \( \boxed{3} \).