5. An integer $n$ is called a "good" number if $|n|$ is not a perfect square. Find all integers $m$ such that $m$ can be expressed as the sum of three distinct "good" numbers in infinitely many ways, and the product of these three "good" numbers is a perfect square of an odd number.
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Assume that $m$ can be expressed as $m=u+v+w$, and $u w w$ is the square of an odd number. Therefore, $u, v, w$ are all odd numbers, and $u w \equiv 1(\bmod 4)$. Thus, either two of $u, v, w$ are congruent to 3 modulo 4, or none of them are. In either case, we have
$$
m=u+v+w \equiv 3(\bmod 4).
$$

Next, we prove that when $m=4 k+3$, the required conditions are satisfied. For this, we seek an expression of the form $4 k+3=x y+y z+z x$. In such an expression, the product of the three addends is a perfect square. Let $x=2 l+1, y=1-2 l$, then we can deduce that $z=2 l^{2}+2 k+1$. Thus, we have
$$
\begin{array}{l}
x y=1-4 l^{2}=f(l), \\
y z=-4 l^{3}+2 l^{2}-(4 k+2) l+2 k+1=g(l), \\
z x=4 l^{3}+2 l^{2}+(4 k+2) l+2 k+1=h(l).
\end{array}
$$

From the above expressions, it is clear that $f(l), g(l), h(l)$ are all odd numbers, and their product is the square of an odd number. It is also easy to see that, except for a finite number of $l$, $f(l), g(l), h(l)$ are distinct.

Next, we prove that for infinitely many $l$, $|f(l)|, |g(l)|, |h(l)|$ are not perfect squares.
When $l \neq 0$, $|f(l)|$ is not a perfect square.
Choose two different primes $p, q$, such that $p>4 k+3, q>4 k+3$. Choose $l$ such that $l$ satisfies
$$
\begin{array}{ll}
1+2 l \equiv 0(\bmod p), & 1+2 l \equiv 0\left(\bmod p^{2}\right), \\
1-2 l \equiv 0(\bmod q), & 1-2 l \equiv 0\left(\bmod q^{2}\right),
\end{array}
$$

By the Chinese Remainder Theorem, such an $l$ exists.
Since $p>4 k+3$, and
$$
\begin{array}{l}
2\left(2 l^{2}+2 k+1\right)=(2 l+1)(2 l-1)+4 k+3 \\
\equiv 4 k+3(\bmod p),
\end{array}
$$

Therefore, $2\left(2 l^{2}+2 k+1\right)$ cannot be divisible by $p$.
Thus, $2 l^{2}+2 k+1$ cannot be divisible by $p$.
Hence, $|h(l)|=\left|(2 l+1)\left(2 l^{2}+2 k+1\right)\right|$ is divisible by $p$ but not by $p^{2}$.
Therefore, $|h(l)|$ is not a perfect square.
Similarly, we can show that $|g(l)|$ is also not a perfect square.