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Im looking for non-UFD rings such that factoring of any element of that ring into irreducibles leads to either all factorizations squarefree or all factorizations squareful.

Thus let $n$ be an element of that ring then either $a^2 b c ... = A^2 B C ...= $ etc where the variables are distinct irreducibles. Or we get $x y z ... = X Y Z ...= etc$ with the variables distinct irreducibles. But we cannot get $a^2 b c ... = s d f ...$

This means if we have factorization of $n$ where at least one irreducible element divides $n$ twice than all factorizations are squareful. And if we have a factorization of $n$ where no irreducible element divides $n$ twice than any other factorization of $n$ lacks an irreducible than devides $n$ twice.

UFD -> http://en.wikipedia.org/wiki/Unique_factorization_domain

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To be clear, the question is asking for an example of a commutative Noetherian domain R, which is not a UFD, such that if $a_1^{k_1} \cdots a_l^{k_l}=b_1^{m_1} \cdots b_p^{m_p}$ for distinct irreducibles $a_1,\dots,a_k \in R$ and distinct irreducibles $b_1,\dots,b_p \in R$, then if some $k_i>1$ we must also have some $m_j>1$? –  S123 Nov 1 '12 at 15:11
    
Seems unlikely there'd be such a thing. Note that if $ab=cd$ and $ae=fg$ then $a^2be=cdfg$. –  Gerry Myerson Nov 2 '12 at 12:14
    
@GerryMyerson : I think I am even willing to accept that as a proof. If you want to answer. –  mick Nov 3 '12 at 22:19
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up vote 1 down vote accepted

Elevating my comment to an answer, per suggestion of OP:

If $m=ab=cd$ and $n=ae=fg$ then $mn=a^2be=cdfg$, so it seems unlikely that there is a ring with the property requested.

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