3. Determine all natural numbers $n$ and all prime numbers $p$ such that

$$
17^{n} \cdot 2^{n^{2}}-p=n^{2}\left(2^{n^{2}+3}+2^{n^{2}}-1\right)
$$
The equation will be written in the form

$$
\left(17^{n}-9 n^{2}\right) \cdot 2^{n^{2}}+n^{2}=p
$$

For $n=1$ we have $p=17$ and this is a prime number, so the pair $(n, p)=(1,17)$ is one solution to the problem.

Let $n>1$. By mathematical induction, it is easily proven that in this case $17^{n}-9 n^{2} \geq 1$, so $\left(17^{n}-9 n^{2}\right) \cdot 2^{n^{2}}+n^{2}>2$, i.e., $p>2$. Therefore, $p=\left(17^{n}-9 n^{2}\right) \cdot 2^{n^{2}}+n^{2}$ is an odd number, which means that $n$ is an odd number. Hence, $n^{2}=4 k(k+1)+1=8 t+1$, for some $t \in \mathbb{N}$. Now, if we use the fact that $2^{8} \equiv 1(\bmod 17)$, we get

$$
\begin{aligned}
\left(17^{n}-9 n^{2}\right) \cdot 2^{n^{2}}+n^{2} & \equiv(-9) n^{2} \cdot 2^{n^{2}}+n^{2} \\
& =(-9) n^{2} \cdot 2^{8 t+1}+n^{2} \\
& =(-18) n^{2} \cdot\left(2^{8}\right)^{t}+n^{2} \\
& \equiv(-18) n^{2}+n^{2} \\
& =-17 n^{2} \equiv 0(\bmod 17)
\end{aligned}
$$

Therefore, for $n>1$ the left side of (1) is divisible by 17, so since $p$ is a prime number, we conclude that $p=17$. However, then for $n \geq 2$ we get

$$
\left(17^{n}-9 n^{2}\right) \cdot 2^{n^{2}}+n^{2} \geq 1 \cdot 2^{4}+4=20>17=p
$$

which is a contradiction. Finally, from the obtained contradiction, it follows that there is no solution such that $n>1$, i.e., $(n, p)=(1,17)$ is the only solution to the problem.