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Find all integers $k>2$ such that $5\equiv k \bmod k^2$.

I ended up with quardratic formula. Is it right?

2 Answers 2

8

$k^2|k-5$

so $k|k-5$

so

$k|5$

And since we stipulate $k>2$, we only need check $k=5$.

6

If $k>5$, then $k^2 > k - 5 > 0$, so $k-5$ can't be divisible by $k^2$; so the proposition is false. Since you were asked for integers $>2$, the only possibilities are 3, 4 and 5. A quick check reveals that 3 and 4 don't work, but 5 does; so the only possible value is $k=5$.

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    @Gigili: Because David wants to eliminate all values of $k$ greater than $5$ by showing that they can’t possibly be solutions to the congruence. Once they’ve been eliminated, he can go on to consider the other possible values of $k$ individually. Since there are only three of them, $3,4$, and $5$, this is easy.2012-01-30