@TobiasKildetoft It can't equal 5. Because if $p$ is prime and therefore odd (except if it equals 2 which it clearly wouldnt) then $p - 1$ is even and i can't multiply any integer by $2$ to get $5$
@TobiasKildetoft I meant, to find the $n$ where $\varphi(n) = 2$ is either $n = 3$ or $n = 4$
@Kaish it might, but we would need the other factor to give $5$ which we saw was not possible
in fact, $\varphi(n)$ is even unless $n$ is $1$ or $2$ (my favorite proof is: $-1$ has order $2$ in the units mod $n$ except in those degenerate cases)
@TobiasKildetoft OHHHHHH. And, then, just because its seems like a clever idea, we look at the next number above $10$ and notice it is $11$ and this is prime and so here, we can set $m = 0$. And so we see that $k = 11$ is odd and so we get another one where $2k =22$. For this, we can set $m = 1$
@TobiasKildetoft How would we know that these are all the possibilities?
@Kaish well, if $(p-1)p^m = 10$ then $p-1$ is a divisor of $10$, so it is either $1$, $2$, $5$ or $10$. But we can easily check that no $m$ will work exceot in the case of $p = 11$
@TobiasKildetoft Ohh, ok. Thank you very much. Lol I really like this Number theory module but its really annoying that these things make sense and everything starts getting interestings only when its exam period and so I can't really dwel on them much
But likely they require a proficiency in a more-or-less common tag that isn't already camped by more cunning entities. No such tags exist for me... yet.
First one proves that a system of the form $$x\equiv \delta_{ik}\mod m_k\; ;\; 1\leq k\leq r$$ has unique solution for each choice of $i=1,\dots,r$. The proof is simply by induction, and can be proven WLOG for the case $i=1$. Then, to prove the general case, one simply takes $$x=\sum_{i=1}^r a_i x_i$$ where the $x_i$ are the solution to the $\delta_{ik}$ systems and the $a_i$ are what we want.
(by the $\delta_{ik}$ I mean $x\equiv 1 \mod m_1,x\equiv 0\mod m_2,\equiv 0\mod m_3,\dots$)
CRT says that each $(a_1,\dots,a_n)$ has one and only one corresponding element in ${\bf Z}/m\bf Z$ and the iso is precisely $(a_1,\dots,a_n)\mapsto \sum a_ix_i$
Perhaps someone could kick you right in the kisser unsollicited. Then tell you you should be thankful. How would that be any different? Only the reason that most people dislike it rather than like?
