4
$\begingroup$

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

I ended up with quardratic formula. Is it right?

$\endgroup$
0

2 Answers 2

8
$\begingroup$

$$k^2|k-5$$

so $$k|k-5$$

so

$$k|5$$

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

$\endgroup$
6
$\begingroup$

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 $k=5$.

$\endgroup$
3
  • 1
    $\begingroup$ Why $k>5$ while it's true for $k>2$? $\endgroup$
    – Gigili
    Commented Jan 30, 2012 at 7:12
  • 3
    $\begingroup$ Because I needed $k>5$ for the main part of the proof - if I don't know $k-5>0$ I can't show that it's not a multiple of $k^2$. That's why I dealt with $k=3,4,5$ as special cases. $\endgroup$
    – user22805
    Commented Jan 30, 2012 at 8:11
  • $\begingroup$ @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. $\endgroup$ Commented Jan 30, 2012 at 9:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .