Given the equations:
$$p^2 + qt = (p + t)^n$$
$$p^2 + qr = t^4$$

We need to show that $n < 3$ and determine all the numbers $p, q, r, t, n$ that satisfy the given conditions.

### Part (a): Show that $n < 3$

1. Assume $n \geq 3$. Then:
$$(p + t)^n \geq (p + t)^3$$

2. Expanding $(p + t)^3$ using the binomial theorem:
$$(p + t)^3 = p^3 + 3p^2t + 3pt^2 + t^3$$

3. Since $p^2 + qt = (p + t)^n$, we have:
$$p^2 + qt \leq (p + t)^n$$

4. For $n \geq 3$:
$$(p + t)^n \geq p^3 + 3p^2t + 3pt^2 + t^3$$

5. Comparing the terms:
$$p^2 + qt \leq p^3 + 3p^2t + 3pt^2 + t^3$$

6. Since $qt \leq 3p^2t + 3pt^2 + t^3$, and $qt$ is linear in $t$ while the right-hand side is cubic in $t$, it is clear that:
$$p^2 + qt < p^3 + 3p^2t + 3pt^2 + t^3$$

7. This leads to a contradiction because $p^2 + qt$ cannot be greater than or equal to $(p + t)^n$ for $n \geq 3$.

8. Therefore, $n < 3$.

### Part (b): Determine all the numbers $p, q, r, t, n$

1. From part (a), we know $n \leq 2$. Since $p^2 + qt > p + t$, $n$ cannot be 1. Thus, $n = 2$.

2. Given $n = 2$:
$$p^2 + qt = (p + t)^2$$
$$p^2 + qt = p^2 + 2pt + t^2$$

3. Simplifying:
$$qt = 2pt + t^2$$
$$t(q - 2p) = t^2$$
$$t(q - 2p - t) = 0$$

4. Since $t \neq 0$:
$$q - 2p = t$$

5. Substituting $t = q - 2p$ into the second equation:
$$p^2 + qr = (q - 2p)^4$$

6. Simplifying modulo $q$:
$$p^2 + qr \equiv 0 \pmod{q}$$
$$(q - 2p)^4 \equiv 0 \pmod{q}$$

7. Since $q$ is a prime number:
$$q \mid (q - 2p)^4$$
$$q \mid 16p^4 - p^2$$
$$q \mid p^2(4p - 1)(4p + 1)$$

8. Since $q > 2p$:
$$q \mid (4p - 1)(4p + 1)$$

9. Considering $q \geq 4p - 1$ and $q \geq 4p + 1$:
$$q = 4p - 1 \text{ or } q = 4p + 1$$

10. For $q = 4p - 1$:
$$p^2 + (4p - 1)r = (2p - 1)^4$$
$$r = \frac{(2p - 1)^4 - p^2}{4p - 1}$$

11. For $p = 2$:
$$q = 4(2) - 1 = 7$$
$$t = q - 2p = 7 - 4 = 3$$
$$r = \frac{(2(2) - 1)^4 - 2^2}{4(2) - 1} = \frac{3^4 - 4}{7} = \frac{81 - 4}{7} = 11$$

12. Thus, $(p, q, r, t, n) = (2, 7, 11, 3, 2)$.

13. For $q = 4p + 1$:
$$r = \frac{(2p + 1)^4 - p^2}{4p + 1}$$
$$r = (p + 1)(4p^2 + 3p + 1)$$

14. Since $r$ must be prime, $4p^2 + 3p + 1$ must be 1, which is not possible for any prime $p$.

The final answer is $\boxed{ (p, q, r, t, n) = (2, 7, 11, 3, 2) }$