$$\begin{array}{l} \text { 3. Let } 2 \equiv 2(\bmod 5), 2^{2} \equiv 4(\bmod 5) \text {, } \\ 2^{3} \equiv 3(\bmod 5), 2^{4} \equiv 1(\bmod 5), \\ \text { then } 2^{\mathrm{D}} \equiv 2^{\mathrm{n}+4}(\bmod 5) \text { . } \\ \end{array}$$

Similarly, $3^{\mathrm{n}} \equiv 3^{\mathrm{n}+4}(\bmod 5)$,
$$4^{n} \equiv 4^{n+4}(\bmod 5) \text {. }$$

Therefore, $p_{\mathrm{a}} \equiv p_{n+{ }_{4}}(\bmod 5)$.
Also, $p_{1} \equiv 0(\bmod 5), p_{2} \equiv 0(\bmod 5)$,
$$p_{3} \equiv 0(\bmod 5), p_{4} \equiv 4(\bmod 5),$$

Thus, except for multiples of 4, for all $n$, $p_{0} \equiv 0(\bmod 5)$, which means the answer is all positive integers that are not multiples of 4.
$$$$