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Determine for which $n \in \mathbb N^*$ values the following equations in $\mathbb Z_n$ are satisfied:

  • $[3] \cdot [5] = [3] + [5]$
  • $[3] \cdot [5] = [27]$ and $[3] + [5] = [11]$

I can't come up with anything leading to a solution, so every piece of advice is very welcome.

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In other words, you want $n$ such that $15\equiv 8$ in the first case and $15\equiv 27$ and $8\equiv 11$ mod $n$. – lhf Sep 3 '12 at 10:44
Use the definitions of $\cdot$, $+$, $[n]$ and $=$. You get conditions that $n$ must satisfy. Work with them. – Jyrki Lahtonen Sep 3 '12 at 10:44
up vote 1 down vote accepted

Without the bothering square parentheses, just taking into account we operate modulo $\,n\,$ : $$3\cdot 5=3+5\Longleftrightarrow 3\cdot 5-3=5\Longleftrightarrow 3\cdot 4=5\Longleftrightarrow 12=5\pmod n$$ $$3\cdot 5=27\,\,\wedge\,\,3+5=11\Longleftrightarrow 15=27\pmod n\,\,\wedge\,\,8=11\pmod n$$

So for the first case above, we need $\,12=5\Longleftrightarrow 7=0\pmod n\,$ , so not many options here...

In the second case, $\,12=27\pmod n\Longleftrightarrow 0=15\pmod n\,$, and also $\,8=11\pmod n\Longleftrightarrow 0=3\pmod n\,$

The second condition leaves us no option, and this option also fulfills the first one, so we're done.

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Did you really want to complete this homework assignment? – Gerry Myerson Sep 3 '12 at 12:42
Yeah, why not? After this I hope he can extrapolate what he might have learnt with this one and apply it to other questions. If he continues asking the same things over and over, as many others around here then I shall not help him out. I tend to indulge in this kind of questions as they're classical from the very first courses in abstract algebra and can be tough to grasp. – DonAntonio Sep 3 '12 at 13:04
In the first place, OP asked for advice, not solutions. And in the second place, OP might have learned more from seeing a partial solution and then figuring out the rest. – Gerry Myerson Sep 3 '12 at 13:16
That's not true: he wrote "every piece of advice is very welcome", and didn't rule out anything else. What he could have learned more from is a matter of guessing since we don't know the OP. – DonAntonio Sep 3 '12 at 14:21
First of all thank you for your help, I didn't ask for any solution I sort of managed to solve my exercise by reading hfl's hint, which was precious. I actually asked for an advice, but, after all, personally speaking, I am glad to have a complete answer to compare my calculation with, they make me 100% sure I'm doing ok. :) – haunted85 Sep 3 '12 at 18:05

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