Is there a proof or counterproof of the following statement?
An integer $i\in$ $Z^+$ exists such that $a*b=i$ and $c*d=i$ where $a,b,c,d\in$ $Z^+$ and $a\neq b\neq c\neq d\neq 1$ .
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1Have you spent any time at all thinking about this? – David C. Ullrich Dec 06 '15 at 17:28
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5$4\cdot3=2\cdot6$ or is there something I'm missing? – egreg Dec 06 '15 at 17:28
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@DavidC.Ullrich OP is asking is there exists a proof or counterproof, not for either one – BCLC Dec 06 '15 at 17:29
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Looking at the title I think he/she might mean this : It is possible that $ab=cd$ , $ac=bd$ and $ad=bc$ where $a,b,c,d$ are all distinct ? – Dec 06 '15 at 17:31
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You seem to be asking whether there exists a positive integer $n$ which can be written $n=ab$ and $n=cd$, where $a,b,c,d$ are pairwise distinct integers greater than $1$.
Yes: $12=2\cdot6=3\cdot4$
If you want to know if there exist $a,b,c,d$ pairwise distinct integers greater than $1$ such that $ab=ac=ad=bc=bd=cd$, then the answer is surely no, because from $ab=ac$ you get $b=c$.
egreg
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Is the distinction between 'pairwise distinct' and 'distinct' the same as the distinction between 'pairwise disjoint' and 'disjoint' ? – BCLC Dec 06 '15 at 17:40
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1@BCLC It's just being picky; most of the times one would say “distinct numbers” to mean that no two are equal, but it's better being on the safe side. – egreg Dec 06 '15 at 17:42
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Yes. The easiest way is to think of unique prime factorizations as containing "atoms" out of which each number "molecule" is made, then 2 distinct factors are sufficient: $a\cdot b^2 = ab \cdot b,$ and we can use 4 to make it squarefree $ab\cdot cd = ac\cdot bd.$
CR Drost
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