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6h
comment Is there a connection between lattices in the sense of orders and lattices in the sense of groups?
I think it is easy to see a word in different places in maths and wonder if there is a reason for it. However sometimes you just have to accept that things are named by "intuitive description". All of the things you mention look like "lattices" in some sense but not necessarily in the exact same sense.
9h
revised Prove result without using Fermat's Last Theorem
edited title
Jun
1
comment Is there a counterexample? $\forall\ p \gt 3 \in \Bbb P, (number\ of\ Quadratic\ Residues\ mod\ kp)=p\ when\ k\in\{2,3\}$
Have you tried the same for $4p$? It should still work since there are exactly two squares mod $4$. However the result will not work for $kp$ if $k>4$ since there are definitely more than $2$ squares mod $k$ ($0,1,4,...$).
Jun
1
comment Is there a counterexample? $\forall\ p \gt 3 \in \Bbb P, (number\ of\ Quadratic\ Residues\ mod\ kp)=p\ when\ k\in\{2,3\}$
Yes, which is what was said in Asvin's answer. It is just a consequence of the CRT.
May
29
answered Is there a counterexample? $\forall\ p \gt 3 \in \Bbb P, (number\ of\ Quadratic\ Residues\ mod\ kp)=p\ when\ k\in\{2,3\}$
May
29
comment Lower bounds on the index of $\mathbf Z[X]/(P)$ in the ring of integers of a number field
If $P$ is a rational polynomial then it doesn't necessarily lie in $\mathbb{Z}[X]$...so what is the quotient $\mathbb{Z}[X]/(P)$?
May
10
comment Non-monic polynomial with roots on the unit circle
Scaling a polynomial by $\alpha\in K^{\times}$ doesn't change its roots...so easiest way is to take a monic example and scale by your favourite $\alpha \neq 1$.
Apr
17
awarded  Enlightened
Apr
17
awarded  Nice Answer
Mar
12
revised primitive roots problem. that integer n can never have exactly 26 primitive roots.
edited tags
Mar
12
comment primitive roots problem. that integer n can never have exactly 26 primitive roots.
Well can you solve $\phi(m) = 26$?
Mar
5
comment Fermat's Theorem on $p = a^2 + b^2$
Ah, I forgot, sorry!
Mar
4
comment Fermat's Theorem on $p = a^2 + b^2$
As you probably know one can solve this problem for any $n$. The key observation is that $p = x^2 + ny^2$ is equivalent to $p$ splitting completely in certain ring class fields attached to the order $\mathbb{Z}[\sqrt{-n}]$. One can calculate these explicitly for small $n$ (and the examples you give in your answer are exactly the defining polynomials). In general one can find these fields by adjoining torsion points of elliptic curves with complex multiplication by $\mathbb{Z}[\sqrt{-n}]$.
Mar
4
comment Fermat's Theorem on $p = a^2 + b^2$
The real point is that for class number one it is easy since there will only be one class of quadratic forms (of the given discriminant) and so representation of primes by such forms is determined by the Legendre symbol alone. When the class number rises you have to discard certain classes determined by the Legendre symbol...and these are determined by the splitting of certain polynomials mod $p$.
Mar
2
comment Can $ 2^{3^{4^{.^{.^{.^{n-1}}}}}}\equiv 1 \bmod {n} $ for some $n>7$.
The title has been edited!
Mar
2
revised Can $ 2^{3^{4^{.^{.^{.^{n-1}}}}}}\equiv 1 \bmod {n} $ for some $n>7$.
edited title
Mar
2
comment Can $ 2^{3^{4^{.^{.^{.^{n-1}}}}}}\equiv 1 \bmod {n} $ for some $n>7$.
Oh my bad...I was reading the title of the question!
Mar
2
answered Can $ 2^{3^{4^{.^{.^{.^{n-1}}}}}}\equiv 1 \bmod {n} $ for some $n>7$.
Mar
1
answered For which integers $x$ does $\varphi(x)$ divide $3^x + 1$?
Mar
1
comment Cubic Residues (Number Theory)
When looking for quadratic residues you usually exclude $p=2$ (because everything here is a quadratic residue). When looking for cubic residues you will have to make a similar condition...since for some primes EVERYTHING is a cubic residue (e.g. $p=3,5,11,...$). The primes that are left over will satisfy results similar to quadratic residues (there will be $\frac{p-1}{3}$ cubic residues, however the multiplicativity will fail but there will still be SOME group structure etc).