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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 $n$.2012-09-03
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    Use the definitions of $\cdot$, $+$, $[n]$ and $=$. You get conditions that $n$ must satisfy. Work with them.2012-09-03

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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?2012-09-03
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    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.2012-09-03
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    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.2012-09-03
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    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.2012-09-03
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    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. :)2012-09-03