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Suppose that there is a set of natural numbers $a, b, c, d...$

Is there any such set that any product of two different numbers is a multiple of $x$, but any product of two same numbers - that is a number squared - is not a multiple of $x$?

How does one determine such set?

Edit: I want to know a general method for determining for any cardinality of a set. Not just two or so.

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Are you another alias for TTTY, who has been asking variations on this question lately? – Gerry Myerson Oct 31 '12 at 4:27
The first part of your question would be clearer if you wrote "Does there exist a subset $S$ of the natural numbers such that any product..." A hint - consider the prime factorizations of members of a hypothetical such set. – Jonah Sinick Oct 31 '12 at 5:34
up vote 3 down vote accepted

To exhibit an $n$-element set with this property, choose $n$ distinct primes $p_1,p_2,\dots,p_n$; then let $x=\left(\prod_i p_i\right)$ and $$ S=\left\{\frac{x}{p_1}, \frac{x}{p_2},\dots,\frac{x}{p_n}\right\}. $$ Each entry is missing one of the prime factors of $x$, and so is any power of a single entry; but the product of any two distinct entries is divisible by $x$.

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What about $\,\{2,3\}\,$? The product of any two different of these numbers is a multiple of $\,6\,$ but none of them squared is such a multiple...

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And for a three element set $$\{2 \cdot 3^4 \cdot 5^7, 2^4 \cdot 3^7 \cdot 5^1, 2^7 \cdot 3^1 \cdot 5^4\}$$ $x = 2^5 3^5 5^5$.

An example in the general case if, for a $n$ element set, first choose $n$ distinct primes, say $p(0),p(1),p(2),\ldots, p(n-1)$, then the set is given by $$S = \left \{ p(\overline{k}), p(\overline{k+1})^4, p(\overline{k+2})^7, \cdots, p(\overline{k+n-1})^{3n-2} : \overline{k} = k \pmod{n}\right \}$$

and $x = (p(0) p(1) p(2) \cdots p(n-1))^5$.

Once you get this idea, you can construct infinite other counterexamples as well.

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