Recent content by PaulRS

The result is true for $a,b,c>0$ in general.

I do not know what you mean by that, in fact, you want $m$ and $n$ to be coprime, not be both multiples of $3$. The proof is rather simple and I am going to spell it out completely so that you see the method to it, let $d$ be the greatest common multiple of $m$ and $n$. Then clearly $d$...

Condition. Observe that $$ m + n = 2\,x + 1\,,\qquad m-n=2y+1\,. $$ Under your hypothesis, I suggest you prove that $$ \gcd(m,n)=\gcd(m+n,m-n) = \gcd(2x+1,2y+1)\,. $$ Here the fact that $m+n$ and $m-n$ are odd (well, that at least one of them is odd) will be crucial. Example. When $x=10$ you...

Hi mathmari! Yes it is correct, but that expansion is not that useful because we seek a multiplicative property. Also, the fact that $t^{-1}$ can be used won't ever be of great help (try to justify why), and thus we shall assume that $a$ is a polynomial in $t$. Converse. Try a few small cases...

Yes, you are right indeed, it follows from $ F_{k\,n} = \sum_{j=1}^k \binom{k}{j} F_j\, F_n^j F_{n-1}^{k-j} $ and inductive reasoning. Simply pick $k := a$, $ n := a^{d-1}\,n$ in that equation above, then observe that $F_{a^{d-1}\,n}^j$ is a multiple of $a^{d+1}$ for $j\geq 2$ (use the...

First, let us forget about the sign and work with unsigned Stirling numbers (so on the left you'd have $\left[{n\atop 3}\right]$, and on the right the same except that the factor $(-1)^{n-1}$ vanishes) . The result then will follow immediately. Now, we have $\left[{n\atop 3}\right] =...

That's strange indeed. If all of these were well-defined, then it follows trivially that $\Phi_m$ is irreducible (and viceversa). So one would expect this proof to be at least as hard as the proof that $\Phi_m$ is irreducible. For all the homomorphisms to be well-defined we must have that...

I have not been able to see page 72 in google books. I am pretty sure that those symbols stand for the product measure though, and that by $dt$ they mean the Lebesgue measure. You should be able to find a section on the product measure in most Real Analysis books (almost everywhere :D).

Another approach: If $n>2$ , then the elements in $\{1,\ldots,n\}$ which are coprime to $n$ come in pairs of distincts elements $\{x,n-x\}$ where $x \leq n / 2$ and $x$ is coprime to $n$. This follows from the fact that $x$ is coprime to $n$ if and only if so is $n-x$. Note that $x \leq n-x$...

Correct. :) There is a simpler polynomial $p(y)\in F_2[y]$ such that $p(x) = 0$, can you find it?

Right. Otherwise $x\in F_2$ and so $x = f(x^2) / g(x^2)$ for some polynomials $f$ and $g$ ($g\neq 0$), which you can check is nonsense.

Here is an idea: Let $F = {\mathbb Q}$ (to fix ideas). Consider $F_1 = F(x)$ and $F_2 = F(x^2)$, and finally $K = F(x)$. The field $F_1$ should be isomorphic to $F_2$ since $x^2$ is trascendental over $F$. Clearly $K/F_1$ has degree 1. However ;).