Find all integers $k \geq 2$ such that $k^2 = 5k(\mod 15).$

Using arithmetic on $\mathbb{Z}_{15},$ $\bar{k}^2 = \bar{5}\bar{k},$ may I divide both sides by $\bar{k}$ to arrive at $\bar{k} = \bar{5}?$

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    $\begingroup$ Given the equation $x^2=5x$ can you divide both sides by $x$ to arrive at $x=5$? $\endgroup$ – Lord Shark the Unknown Oct 8 at 17:36
  • $\begingroup$ if you assume $x \neq 0.$ The answers to that equation are $x = 0, 5.$ Im just not very comfortable with doing arithmetic on modulo. I just read about it last night and am now attempting problems. So, I don't feel very confident with it. I saw no examples of division in the chapter. Are you allowed to divide residual classes like integers? $\endgroup$ – Rafael Vergnaud Oct 8 at 17:38

Only if $k$ has an inverse $m$, with $km=1$, so you multiply by the inverse.
Numbers that are multiples of 3 or 5 don't have inverses. That is because any multiple of them is still a multiple of 3 or 5, and so not 1.
So check multiples of 3 and 5 separately.
Note that if the equation were $k^2=2k$, solutions would be 0,2,5 and 12.

  • $\begingroup$ I assumed that $\bar{k} \neq \bar{3}, \bar{5}.$ Then, I can assume that $\bar{k}$ has an inverse. Hence, $\bar{k}^2 = \bar{5}\bar{k}$ implies $\bar{k} = \bar{5},$ which is contradictory. Did I do something wrong here? $\endgroup$ – Rafael Vergnaud Oct 8 at 18:48

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