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22s
revised Division problems
edited tags
10m
comment the unit group of an infinite field cannot be cyclic
Check out in particular Blue's answer in that older thread, where this problem was discussed.
2h
revised Find the value of the question below
rolled back to a previous revision
6h
comment $n$-vertex $3$-edge-colored graphs with exactly $6$ automorphisms which preserve edge color classes, but permute the edge colors distinctly?
You get a lot more graphs with the following recipe: Triangulate the plane with equilateral triangles. Color the edges parallel to $x$-axis green, the rising edges orange and the descending edges blue. Pick a subset stable under the action of the symmetries of the triangle. The sets will have varying sizes depending on whether the origin is at a vertex or at a centroid of one of the triangles.
6h
revised $n$-vertex $3$-edge-colored graphs with exactly $6$ automorphisms which preserve edge color classes, but permute the edge colors distinctly?
added 38 characters in body
6h
comment $n$-vertex $3$-edge-colored graphs with exactly $6$ automorphisms which preserve edge color classes, but permute the edge colors distinctly?
@Rebecca: I think I thought of a fix for the case $3\mid n$.
6h
revised $n$-vertex $3$-edge-colored graphs with exactly $6$ automorphisms which preserve edge color classes, but permute the edge colors distinctly?
added 38 characters in body
6h
revised $n$-vertex $3$-edge-colored graphs with exactly $6$ automorphisms which preserve edge color classes, but permute the edge colors distinctly?
added 38 characters in body
6h
comment $n$-vertex $3$-edge-colored graphs with exactly $6$ automorphisms which preserve edge color classes, but permute the edge colors distinctly?
You're right. Sorry.
6h
comment $n$-vertex $3$-edge-colored graphs with exactly $6$ automorphisms which preserve edge color classes, but permute the edge colors distinctly?
It might be more interesting if you add more requirements. I don't know what would be useful for the purposes of your goal of applying this to Latin rectangles.
6h
answered $n$-vertex $3$-edge-colored graphs with exactly $6$ automorphisms which preserve edge color classes, but permute the edge colors distinctly?
7h
comment Find the value of the question below
For questions of this type the key is to observe that the truly interesting variable is $x+\dfrac1x$ - not $x$ itself. +1 for making that observation.
7h
comment Find the value of the question below
In case you need a confirmation, here's the word from your friendly moderator. DO NOT REPLACE AN EXISTING QUESTION WITH ANOTHER ONE. This is considered rude to the people who answered the original version, because you make their posts look out of place. Remember, the answers are not just for you - they are (mostly ?) for the benefit of all the future readers of this thread.
15h
comment Cyclotomic extension of $K$, $Gal_{K}{F}$ is isomorphic to a subgroup of $\mathbb Z_n^*$
Good. Have you seen a proof of $$\cos\frac{2\pi}5=\frac{-1+\sqrt{5}}4?$$ A more general result is that for a prime $p>2$ the unique quadratic subfield of $\Bbb{Q}(\zeta_p)$ is $\Bbb{Q}(\sqrt{p^*})$, where $p^*=p$, if $p\equiv1\pmod4$ and $p^*=-p$, if $p\equiv -1\pmod4$. Here is my proof for the fact that $\sqrt{-7}\in\Bbb{Q}(\zeta_7)$. One uses Gauss' sums to prove this for a general prime $p$.
16h
answered Cyclotomic extension of $K$, $Gal_{K}{F}$ is isomorphic to a subgroup of $\mathbb Z_n^*$
16h
comment Character sum of a type of “almost linear” surjective mappings over finite fields
Another possibility is that when $\gcd(m,q-1)$, and $c$ ranges over $F^*$ the elements $f(cx)=c^mf(x)$ do the same unless they are all zero. This gives us a lot of cancellation in the sum, but there may be problems with those zeros (also because the sum over $F^*$ gives a non-cancelled $-1$.
16h
comment Character sum of a type of “almost linear” surjective mappings over finite fields
Oops. Yes, that's what I was thinking, but writing... Thinking about the rest. What kind of polynomials did you check?
22h
comment Polynomial Interpolation When $y_i$'s are Permuted
The answer may depend a little bit on the field the entries come from. Assuming that they are reals (though data integrity application suggests otherwise). If we fix $(x_1,x_2,\ldots,x_n)$ and generate real numbers $(y_1,y_2,\ldots,y_n)$ with some common random process (uniform from some interval, i.i.d. gaussian,...) then the probability for the interpolation polynomial to have degree $<n-1$ is zero. Checking out the finitely many ($n!$) permutations is not going to change that, still $P=0$.
1d
answered The topology of $\mathbb{Z}_p$
1d
comment Polynomials with range containing an arithmetic progression
LOL! A good one, @Gerry.