Demonstrate that there are no positive integers $a, b, c, d$, relatively prime, such that $ab + cd$, $ac + bd$, and $ad + bc$ are odd divisors of
$$
(a + b - c - d) (a - b + c - d) (a - b - c + d)
$$
Try the following: Being a prime number, the integer $\beta$ numbers that divide $\alpha^n$ ($\beta | \alpha^n$) are multiples of $\alpha$. $\beta = \alpha^k$, $n \ge k \ge 0$. Example: $n|7^4$ logo $n \in\{1, 7, 7^2, 7^3, 7^4\}$.
It's a matter of the '' Southern Cone ''. I don't remember the year
I don't see how your observation of the divisors of powers of primes can be applied here. The question says $a,b,c,d$ are relatively prime, which means no $2$ of them have any factor in common, but none of them are necessarily prime and, in fact, one or more may be $1$, not to mention possibly being composite numbers.
Instead, first note $a,b,c,d$ are each relatively prime to the $3$ odd numbers you're asking about, i.e.,
$$e = ab + cd \tag{1}\label{eq1A}$$
$$f = ac + bd \tag{2}\label{eq2A}$$
$$g = ad + bc \tag{3}\label{eq3A}$$
This is because each of $e,f,g$ uses all $4$ of $a,b,c,d$ once in one of the $2$ terms. For example, if $\gcd(a,e) = h$, then $h \mid a$ so $h \mid ab$, plus $h \mid cd$ as well, which can only be true if $h = 1$.
Since each of $e,f,g$ are odd, this means one of $a,b,c,d$ is even and the other $3$ odd. Next, have
$$i = a + b - c - d = (a + b) - (c + d) \tag{4}\label{eq4A}$$
$$j = a - b + c - d = (a + c) - (b + d) \tag{5}\label{eq5A}$$
$$k = a - b - c + d = (a + d) - (b + c) \tag{6}\label{eq6A}$$
Since each of $i,j,k$ is odd, this also means the value to check of
$$m = (a + b - c - d) (a - b + c - d) (a - b - c + d) = ijk \tag{7}\label{eq7A}$$
is odd as well, in particular it can't be $0$.
Note the values of $e,f,g$ and $i,j,k$ are basically symmetric wrt $a,b,c,d$. This is because there are $\binom{4}{2} = 6$ ways to choose $2$ values out of $a,b,c,d$. Pairing any given $2$ of these with the remaining $2$ variables gives $3$ possibilities. Multiplying each the $2$ chosen values with themselves & adding the results gives $e,f,g$ in some order. Adding each of these $2$ values and subtracting the second set from the first gives, apart from possibly a difference in sign, $i,j,k$ in some order. This symmetry means that basically anything found about a factor, or lack of one, for one case will also apply to the other similar cases.
Next, consider $n = \gcd(e,f)$. If $n \gt 1$, then let $p$ be an odd prime with $p \mid n$. Since $n \mid e - f$, then $p \mid e - f$, i.e., you have
$$\begin{equation}\begin{aligned} ab + cd - ac - bd & \equiv 0 \pmod p \\ a(b-c) + d(c-b) & \equiv 0 \pmod p \\ (a-d)(b-c) & \equiv 0 \pmod p \end{aligned}\end{equation}\tag{8}\label{eq8A}$$
Thus, $a \equiv d \pmod p$ and/or $b \equiv c \pmod p$. Consider first $a \equiv d \pmod p$. For $p \mid m$ requires $p$ divides at least one of $i,j,k$. With $i \equiv (a - d) + (b - c) \equiv b - c \equiv 0 \pmod p$, you get $b \equiv c$. However, this means $e \equiv ab + cd \equiv ab + ba \equiv 2ba \equiv 0 \pmod p$. Thus, $p \mid a$ or $p \mid b$ but, as noted earlier, this is not possible. You get a similar result for $j$. For checking $k$, first note $f \equiv ac + bd \equiv ac + ba \equiv a(c+b) \equiv 0 \pmod p$, so $b + c \equiv 0 \pmod p$. Thus, you have $k \equiv a - (b + c) + d \equiv 2a \equiv 0 \pmod p$, which is not possible. Thus, $p$ doesn't divide any of $i,j,k$, with the same conclusion occurring if you repeat this procedure for the case of $b \equiv c \pmod p$. This means the original assumption of there being any common factors between $e$ and $f$ must be false, i.e., $e$ and $f$ are relatively prime.
You can use similar arguments to show $e$ and $g$, plus $f$ and $g$, are also all relatively prime. Thus, for $e,f,g$ to all be factors of $m$ requires $efg \mid m$. However, since $a,b,c,d$ are all positive integers, you have that
$$e = ab + cd \ge \max(a+d,b+c) \gt |(a + d) - (b + c)| = |k| \tag{9}\label{eq9A}$$
$$f = ac + bd \ge \max(a+b,c+d) \gt |(a + b) - (c + d)| = |i| \tag{10}\label{eq10A}$$
$$g = ad + bc \ge \max(a+c,b+d) \gt |(a + c) - (b + d)| = |j| \tag{11}\label{eq11A}$$
Multiplying \eqref{eq9A}, \eqref{eq10A} and \eqref{eq11A} together gives
$$efg \gt |ijk| = |m| \tag{12}\label{eq12A}$$
Since $m \neq 0$, this shows $efg \not\mid m$, i.e., not all of $e,f,g$ can be divisors of $m$.