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Suppose there is a set of unequal natural numbers. The cardinality of the set is $n$.

For each number, $a_n$, in the set, ${a_n}^2$ is always nonzero multiples of $x$. ($x$ is nonzero integer.)

The product of all numbers in the set never becomes nonzero multiples of $x$.

How does one construct such set?

Also, if one replaces the constraint natural number/integer with some mathematical objects, while retaining "nonzero (integer) multiples of $x$", how does one construct such set?

2 Answers 2

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If $a^2$ is a multiple of $x$, and $b^2$ is a multiple of $x$, then $ab$ is a multiple of $x$. So there is no such set with more than 1 element.

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"if one replaces the constraint natural number/integer with some mathematical objects" then there are nontrivial examples. For example, $\pmatrix{1&3\cr5&7\cr}^2=\pmatrix{1&3\cr5&8\cr}\pmatrix{8&0\cr0&8\cr}$ and $\pmatrix{2&1\cr4&6\cr}^2=\pmatrix{1&1\cr4&5\cr}\pmatrix{8&0\cr0&8\cr}$ but $\pmatrix{1&3\cr5&7\cr}\pmatrix{2&1\cr4&6\cr}=\pmatrix{14&19\cr38&47\cr}$ is not an integer matrix times $\pmatrix{8&0\cr0&8\cr}$.

More abstractly, let $F$ be any field, and consider the ring $F[a,b,c,d,e]/(a^2-ce,b^2-de)$ In this ring, $a^2=ce$, and $b^2=de$, but $ab$ is not a multiple of $e$.