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!
try to see the meaning of the congruence
$\displaystyle 11\equiv k^2(\mbox{mod}\;k)$ means
$\displaystyle \exists n\in \mathbb{Z}(11-k^2=kn)$
$\displaystyle \Rightarrow n\in \mathbb{Z}\;\mbox{and}\;11-k^2=kn$
$\displaystyle \Rightarrow n=\frac{11-k^2}{k}=\frac{11}{k}-k$
since k is already an integer >= 3 , and n must be an integer. the only way n could be integer is if $\displaystyle \frac{11}{k}$ is also an integer. there is only one value of
k for which this happens........
apply similar reasoning to b)