# mathstudent

I am currently a 1st year Advanced Mathematics University Student. I very much appreciate all the kind people in the stack exchange community and thank you all in advance for sharing your knowledge and expertise with me.

 Nov 12 comment Suppose that $n$ is a positive integer with $\phi(n)=8$. Show that if $p$ is an odd prime divisor of $n$ then $p=3$ or $p=5$ahh yes. Many thanks! Nov 12 asked Suppose that $n$ is a positive integer with $\phi(n)=8$. Show that if $p$ is an odd prime divisor of $n$ then $p=3$ or $p=5$ Nov 9 comment $n$ is a positive integer and let $p$ be a prime divisor of $n^{54}+n^{27}+1$. Prove if $p \ne 3$ then $ord_p(n) = 81$ and $p \equiv 1 \pmod {81}$ahh, I understand now. The only thing I would like to clear up is what is meant by - $x-1$ is a unit mod $p$? I'm guessing that it's just $1$ in mod p? And how does that follow from $x \not\equiv 1 \pmod p$? The fraction kind of confuses me, but besides that I do follow your solution. Many thanks! Nov 9 comment $n$ is a positive integer and let $p$ be a prime divisor of $n^{54}+n^{27}+1$. Prove if $p \ne 3$ then $ord_p(n) = 81$ and $p \equiv 1 \pmod {81}$Thanks anon. Following your directions I need to show that $n^{27} \not\equiv 1 \pmod p$, since $p \ne 3$, but I don't see why this would be the case? Also, I do see that $n^{27 \cdot 3} \equiv 1$, but how can I be sure that this is the order? Nov 9 asked $n$ is a positive integer and let $p$ be a prime divisor of $n^{54}+n^{27}+1$. Prove if $p \ne 3$ then $ord_p(n) = 81$ and $p \equiv 1 \pmod {81}$ Nov 9 accepted Show that if $R_n$ is prime then $n$ must be prime. Nov 9 comment Show that if $R_n$ is prime then $n$ must be prime.Many Thanks Andre, I think I understand (just had to get my head around the logic a bit!). By the way, how did you get that $10^a-1$ is a proper divisor of $10^{ab}-1$? I'm assuming this is a pretty basic lemma? To be respectful of your time, could you just point me to where I can read about it? Many thanks again! Nov 9 comment Show that if $R_n$ is prime then $n$ must be prime.Hi Andre, if I am following your solution correctly, this shows that if $n$ is composite then $R_n$ is composite. So if $n=p$ a prime, I would be left with $10^{p-1}+10^{p-2}+...+10+1$, wouldn't I then have to show that this is then prime? Is there a way to start with assuming $R_n$ is prime and going in that direction? Nov 9 asked Show that if $R_n$ is prime then $n$ must be prime. Nov 2 asked Probability Generating Function of Poisson Distribution Nov 2 accepted Show that for a Geometric distribution, the probability generating function is given by $\frac{ps}{1-qs}$, $q=1-p$ Nov 1 asked Show that for a Geometric distribution, the probability generating function is given by $\frac{ps}{1-qs}$, $q=1-p$ Oct 24 comment Proof using Chinese Remainder Theorem for $n=p_1^{m_1}\cdot p_2^{m_2}\cdots p_k^{m_k}$woops. Sorry for the typos. Oct 24 revised Proof using Chinese Remainder Theorem for $n=p_1^{m_1}\cdot p_2^{m_2}\cdots p_k^{m_k}$edited title Oct 24 asked Proof using Chinese Remainder Theorem for $n=p_1^{m_1}\cdot p_2^{m_2}\cdots p_k^{m_k}$ Oct 24 accepted Prove that for $n=2^k$, $(k \ge 3)$ there are 4 natural numbers less than $n$ that satisfy $b^2 \equiv 9 \pmod n$. Oct 24 comment Prove that there exists only 2 solutions for $x^2 \equiv 9 \pmod {p^k}$, ($p$ an odd prime > 3 and $x$ a natural number < $n$)Oh wow, you have no idea how much you have helped me. That's exactly where I was trying to get to the case of $n$ in general - I knew once I could prove these cases, I could use the CRT to bind them together. Can't thank you enough! Oct 24 revised Prove that for $n=2^k$, $(k \ge 3)$ there are 4 natural numbers less than $n$ that satisfy $b^2 \equiv 9 \pmod n$.added 178 characters in body Oct 24 comment Prove that there exists only 2 solutions for $x^2 \equiv 9 \pmod {p^k}$, ($p$ an odd prime > 3 and $x$ a natural number < $n$)I did actually: math.stackexchange.com/questions/537615/… - Hopefully, can get a good nudge in the right direction soon, but I'll keep working on it! Thanks! Oct 24 accepted Prove that there exists only 2 solutions for $x^2 \equiv 9 \pmod {p^k}$, ($p$ an odd prime > 3 and $x$ a natural number < $n$)