# Modulo division: Find all integers $k \geq 2$ such that $k^2 = 5k(\mod 15).$ [duplicate]

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 $$\bar{k} = \bar{5}?$$

• Given the equation $x^2=5x$ can you divide both sides by $x$ to arrive at $x=5$? – Lord Shark the Unknown Oct 8 at 17:36
• 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? – Rafael Vergnaud Oct 8 at 17:38

Only if $$k$$ has an inverse $$m$$, with $$km=1$$, so you multiply by the inverse.
Note that if the equation were $$k^2=2k$$, solutions would be 0,2,5 and 12.
• 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? – Rafael Vergnaud Oct 8 at 18:48