Problem: Find the largest prime number $p<1000$ for which there exists a complex number $z$ satisfying

Solution: Denote $z = a + i b$ . Thus, $a^2 + b^2 = p$
Thus, \[z^3 = a \left( a^2 - 3 b^2 \right) + i b \left( - b^2 + 3 a^2 \right) .\]
Because $p$ ${\rm Re} \left( z^3 \right)$ ${\rm Im} \left( z^3 \right)$ are three sides of a triangle, we have ${\rm Re} \left( z^3 \right) > 0$ and ${\rm Im} \left( z^3 \right) > 0$ .
Thus, \begin{align*} a \left( a^2 - 3 b^2 \right) & > 0 , \hspace{1cm} (1) \\ b \left( - b^2 + 3 a^2 \right) & > 0. \hspace{1cm} (2) \end{align*}
Because $p$ ${\rm Re} \left( z^3 \right)$ ${\rm Im} \left( z^3 \right)$ are three sides of a triangle, we have the following triangle inequalities: \begin{align*} {\rm Re} \left( z^3 \right) + {\rm Im} \left( z^3 \right) & > p \hspace{1cm} (3) \\ p + {\rm Re} \left( z^3 \right) & > {\rm Im} \left( z^3 \right) \hspace{1cm} (4) \\ p + {\rm Im} \left( z^3 \right) & > {\rm Re} \left( z^3 \right) \hspace{1cm} (5) \end{align*}
We notice that $| z^3 | = p^{3/2}$ , and ${\rm Re} \left( z^3 \right)$ ${\rm Im} \left( z^3 \right)$ , and $| z^3 |$ form a right triangle. Thus, ${\rm Re} z^3 + {\rm Im} z^3 > p^{3/2}$ .
Because $p > 1$ $p^{3/2} > p$ .
Therefore, (3) holds.
Conditions (4) and (5) can be written in the joint form as \[\left| {\rm Re} \left( z^3 \right) - {\rm Im} \left( z^3 \right) \right| < p . \hspace{1cm} (4)\]
We have \begin{align*} {\rm Re} \left( z^3 \right) - {\rm Im} \left( z^3 \right) & = \left( a^3 - 3 a b^2 \right) - \left( - b^3 + 3 a^2 b \right) \\ & = \left( a + b \right) \left( a^2 - 4 ab + b^2 \right) \end{align*} and $p = a^2 + b^2$
Thus, (5) can be written as \[\left| \left( a + b \right) \left( a^2 - 4 ab + b^2 \right)  \right| < a^2 + b^2 . \hspace{1cm} (6)\]
Therefore, we need to jointly solve (1), (2), (6).
From (1) and (2), we have either $a, b >0$ , or $a, b < 0$ .
In (6), by symmetry, without loss of generality, we assume $a, b > 0$
Thus, (1) and (2) are reduced to \[a > \sqrt{3} b . \hspace{1cm} (7)\]
Let $a = \lambda b$ . Plugging this into (6), we get \begin{align*} \left| \left( \left( \lambda - 2 \right)^2 - 3 \right)  \right| < \frac{1}{b} \frac{\lambda^2 + 1}{\lambda + 1} .  \hspace{1cm} (8) \end{align*}
Because $p= a^2 + b^2$ is a prime, $a$ and $b$ are relatively prime.
Therefore, we can use (7), (8), $a^2 + b^2 <1000$ , and $a$ and $b$ are relatively prime to solve the problem.
To facilitate efficient search, we apply the following criteria:
\begin{enumerate*}
\item To satisfy (7) and $a^2 + b^2 < 1000$ , we have $1 \leq b \leq 15$ .
In the outer layer, we search for $b$ in a decreasing order.
In the inner layer, for each given $b$ , we search for $a$ .
\item Given $b$ , we search for $a$ in the range $\sqrt{3} b < a < \sqrt{1000 - b^2}$ .
\item We can prove that for $b \geq 9$ , there is no feasible $a$ .
The proof is as follows.
For $b \geq 9$ , to satisfy $a^2 + b^2 < 1000$ , we have $a \leq 30$ .
Thus, $\sqrt{3} < \lambda \leq \frac{30}{9}$ .
Thus, the R.H.S. of (8) has the following upper bound \begin{align*} \frac{1}{b} \frac{\lambda^2 + 1}{\lambda + 1} & < \frac{1}{b} \frac{\lambda^2 + \lambda}{\lambda + 1} \\ & = \frac{\lambda}{b} \\ & \leq \frac{\frac{30}{9}}{9} \\ & < \frac{10}{27} . \end{align*}
Hence, to satisfy (8), a necessary condition is \begin{align*} \left| \left( \left( \lambda - 2 \right)^2 - 3 \right)  \right| < \frac{10}{27} . \end{align*}
However, this cannot be satisfied for $\sqrt{3} < \lambda \leq \frac{30}{9}$ .
Therefore, there is no feasible solution for $b \geq 9$ .
Therefore, we only need to consider $b \leq 8$
\item We eliminate $a$ that are not relatively prime to $b$
\item We use the following criteria to quickly eliminate $a$ that make $a^2 + b^2$ a composite number.
\item For the remaining $\left( b, a \right)$ , check whether (8) and the condition that $a^2 + b^2$ is prime are both satisfied.
The first feasible solution is $b = 5$ and $a = 18$ .
Thus, $a^2 + b^2 = 349$
\item For the remaining search, given $b$ , we only search for $a \geq \sqrt{349 - b^2}$
Following the above search criteria, we find the final answer as $b = 5$ and $a = 18$ .
Thus, the largest prime $p$ is $p = a^2 + b^2 = \boxed{349}$

Answer:  349