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I simply don't understand the question. Doesn't $\mathbb{Z}/42\mathbb{Z}$ consist of $42$ subsets? (Namely $\{0,42,84,\cdots\}, \{1,43,85,\cdots\}, \cdots, \{41, 83, 125, \cdots\}$) And then how would you square an entire subset?

New to this website so if I did anything wrong please tell me.

jacopoburelli
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J.S.A. Frugte
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    Each element is a subset, but squaring a set means squaring any one representative of that set, which leads to another number, which will lie in (exactly one) another set, which we call as the square of this set. For example, the square of the set containing $8$ is the set containing $64$, which also contains $22$ as the unique number between $0$ and $42$, so we say $8^2 = 22$ in $\mathbb Z \over 42 \mathbb Z$. – Sarvesh Ravichandran Iyer Sep 18 '20 at 10:13
  • Same as in the dupe. – Bill Dubuque Sep 23 '20 at 22:26

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@TeresaLisbon is right. This is a question about modulo $42$ residue classes. It asks how many integers satisfy $0\le x\le 41,\,42|x^2-x$.

J.G.
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You are correct, each element of the ring $\mathbb{Z}/42\mathbb{Z}$ is strictly speaking a "coset" - which is a set as you have written. One conviniently thinks about it however as "remainders under division by $42$". So we can choose a "representative" for each element from the set $\{0, ..., 41 \}$

There are two operations to be aware of in $\mathbb{Z}/42\mathbb{Z}$, addition an multiplication - and they occur as you would expect with the above interperatation. Compute the sum (resp. product) in $\mathbb{Z}$ and compute residue. Therefore e.g., $7^2 = 49 \equiv 7 \pmod{42}$.

To your main question, you could simply iterate all of the elements in $\mathbb{Z}/42\mathbb{Z}$. I'll give a different method.

By the Chinese Remainder Theorem $$\mathbb{Z}/42\mathbb{Z} \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/7\mathbb{Z}$$

Then some $a \in \mathbb{Z}/42\mathbb{Z}$ satisfies $a^2 \equiv a \pmod{42}$ if and only if $a^2 \equiv a \pmod{7}$ and $a^2 \equiv a \pmod{2}$ and $a^2 \equiv a \pmod{3}$. Thus it occurs if and only if $a \equiv 0, 1 \pmod{7}$ and $a \equiv 0, 1 \pmod 3$ and $a \equiv 0, 1 \pmod 2$.

Thus the elements of $\mathbb{Z}/42\mathbb{Z}$ satisfying the relation are $0, 7, 21, 28, 36, 1, 15, 22$

Mummy the turkey
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