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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
 

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