6:54 AM
@TheSimpliFire I could not find out what you meant with your hint , but I found the following surprisingly trivial solution : \$ab=cd\$ implies \$(a+c)b=(b+d)c\$ , hence we have \$(a+c)\mid (b+d)c\$. If we assume \$\gcd(a+c,b+d)=1\$, we get \$a+c\mid c\$ which is impossible. Hence there is a prime \$p\$ dividing both \$a+c\$ and \$b+d\$ , hence dividing \$a+b+c+d\$ with \$a+b+c+d>p\$.

@Peter Nice. My hint was that factoring a+b+c+d=(A+B)(C+D) is always possible since we can take a=AC, b=BD, c=BC, d=AD for instance so the sum cannot be prime

Which is always possible since , like you said , \$A,B,C,D\$ can also be \$1\$.

7:10 AM
After having thought about it, can we actually always find such a representation ? It seems to be the case, but how can we prove it ?

@Peter Of course, it satisfies ab=cd since AC * BD = BC * AD
we are basically writing each integer as a product of two integers which can always be done

That the products are equal is clear , but we cannot just choose \$A,B,C,D\$ arbitary, we must ensure that all the \$4\$ equalities are satisfied.

I'm not sure what is unclear: we choose a,b arbitrarily and write each as the product of two integers (a=AC, b=BD), to obtain c,d we swap one of the factors in a with one of the factors in b (that is, c=BC, d=AD)

Let us say , we choose \$A=a\$ and \$B=b\$ , then \$C=D=1\$ implying \$c=B\$ , \$d=A\$. That is what I mean, we must choose \$A,B,C,D\$ carefully.

Yes exactly