1. ## Congruence relation

Hi, also need help with this one, been trying this for hours..

Find all integers that are k>=3 (bigger or equal to 3)

a) 11 ≡ k^2, mod k
b) 7 ≡ k, mod k^2

Thanks!

2. ## Re: Congruence relation

try to see the meaning of the congruence

$11\equiv k^2(\mbox{mod}\;k)$ means

$\exists n\in \mathbb{Z}(11-k^2=kn)$

$\Rightarrow n\in \mathbb{Z}\;\mbox{and}\;11-k^2=kn$

$\Rightarrow n=\frac{11-k^2}{k}=\frac{11}{k}-k$

since k is already an integer >= 3 , and n must be an integer. the only way n could be integer is if $\frac{11}{k}$ is also an integer. there is only one value of
k for which this happens........

apply similar reasoning to b)