I want to prove the following lemma:   $t^n-1$ divides $t^m-1$ in $F[t, t^{-1}]$ if and only if $n$ divides $m$ in $\mathbb{Z}$.   I have done the following:   $\Leftarrow $ :   $n\mid m \Rightarrow m=kn, k \in \mathbb{Z}$  $t^m-1=t^{kn}-1=(t^n)^k-1=(t^n-1)(t^{n(k-1)}+\dots +1)$   So, $t^n-1\mid t^m-1$.   Is this correct?   How could we show the other direction?
2481
For all $a, m, n \in \mathbb{Z}^+$,   $$\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$$
83629
I am trying to solve the following exercise of Dummit and Foote Book(page # 551).   Let $a&gt;1$ be an integer. Prove for any positive integers $n,d$ that $d$ divides $n$ if and only if $a^d-1$ divides $a^n-1$. Conclude in particular that $\mathbb{F}_{p^d}\subseteq\mathbb{F}_{p^n}$ if and only if $d$ divides $n$.  I did the first part and I know that for all $\alpha\in \mathbb{F}_{p^d}$, $\alpha^{p^d}=\alpha$. How can I apply the first part for the second? Any help is greatly appreciated. Thank you.
16747
I'm having problems with an exercise from Apostol's Introduction to Analytic Number Theory.     Given $x$ and $y$, let $m=ax+by$, $n=cx+dy$, where $ad-bc= \pm 1$. Prove that $(m,n)=(x,y)$.   I've tried to give a proof, but I suspect it's wrong (or at least not very good). I would be very thankful for any hints/help/advice!  My proof:     We observe that since $ad-bc= \pm 1$, $ad=bc \pm 1$, and $(ad,bc)=1$.   Now, $(a,b) \mid m$ and $(c,d) \mid n$ but      $$(a,b) = ((d,1)a,(c,1)b)=(ad,a,bc,b)=(ad,bc,a,b)=((ad,bc),a,b)=(1,a,b)=1.$$      Similarly, we determine $(c,d)=1$. So, $1=(a,b) \mid m$ and $1=(c,d) &gt; \mid n$. But $(x,y)$ also divide $m$ and $n$. Since $(x,y) \geq (a,b)=(c,d)=1$, this implies that $(x,y)=m,n$. Hence   $(m,n)=((x,y),(x,y))=(x,y)$.
10357
How can I prove that prove $n^5 - n$ is divisible by $30$? I took $n^5 - n$ and got $n(n-1)(n+1)(n^2+1)$ Now, $n(n-1)(n+1)$ is divisible by $6$. Next I need to show that $n(n-1)(n+1)(n^2+1)$ is divisible by $5$. My guess is using Fermat's little theorem but I don't know how.
21280
I just want to know if my proof to this question is correct. First, I proved it was true for $n = 4$. $$4!&gt;2^4$$ $$24&gt;16$$ Then, I assumed that it was true for $n=k$. $$k!&gt;2^k$$ Afterwards, I tried to prove it for $n=k+1$. $$(k+1)!&gt;2^{k+1}$$ $$(k+1)k!&gt;(2)(2^k)$$ Since $k!&gt;2^k$, am I allowed to simply do $$(k+1)(2^k)&gt;(2)(2^k)$$ $$(k+1)&gt;2$$ and this is true for all $k\geq3$. Would the last two steps be illegal because I only assumed that $k!&gt;2^k$ is true?
4220
I'm trying to prove that $(ma, mb) = $|$m$|$(a, b)$ , where $(ma, mb)$ is the greatest common divisor between $ma$ and $mb$. My thoughts: If $(ma, mb) = d$ , then $d$|$ma$ and $d$|$mb$ → $d$|$max + mby$ → $d$|$m(ax+by)$. This implies that $d$|$m$ or $d$|$(ax+by)$. This is the same as  $d$|$m$ or $d$|$a$ and $d$|$b$, so $d$|$m$ or $d$|$(a,b)$. This is the same as $d$|$m|$ or $d|(a,b)$, so $d$|$|m|(a,b)$. I don't know what to do.
33322
Let $a$ and $b$ be nonzero integers. Prove that $a$ divides $b$ and $b$ divides $a$ if and only if $a = \pm b$.   Since this is a iff statement, I need to prove it both ways:  $\Rightarrow$ If $a=\pm b$, then  $a$ divides $b$ and $b$ divides $a$;  and  $\Leftarrow$ If $a$ divides $b$ and $b$ divides $a$, then $a=\pm b$.  I tried to prove it but I don't know how to manage the $\pm$ sign, can anyone give me a hit or suggestion to start this proof?  Thanks
1549
I have a hard time formulating proofs.  For this problem, I  can see that if $n$ is equal to $8,$ this statement is true.  $(\mathbb{Z}/8\mathbb{Z})^{\times}$ includes elements:  $1,3,5,7$, and all of these are roots of $1-x^2 \pmod 8.$  And obviously $8$ divides $24.$  But how do I prove this without depending on number calculations and only using theorems?  Help Please?  I need a step by step walk through of how to do this proof and what theorems would be appropriate to use.
17207
The question is this:      Let $f:\mathbb C[x,y]\rightarrow\mathbb C[t]$ be the homomorphism that sends $x\mapsto t+1$ and $y\mapsto t^3-1.$ Determine the kernel $K$ of $f$, and prove that every ideal $I$ of $\mathbb C[x,y]$ that contains $K$ can be generated by two elements.   Solution:  I have shown that $((x-1)^3-y-1)=K.$ However, I am having trouble showing the second part. Any hints??
123953
Can someone please show me how to prove this statement? $$\forall x\in\mathbb N (x &gt; 1\to\exists k\in\mathbb N\exists m \in\mathbb N (m \equiv 1 (\text{mod }2) \wedge x = 2^km))$$ I can only assume $m = 3$ and when $x &gt; 1$.
60504
Let $F$ be a field and let $t$ be trascendental over $F$. Prove that $[F(t):F(t^n)]=n$. Obviously $[F(t):F(t^n)]\le n$ since the polynomial $f(x)=x^n-t^n \in F(t^n)$ has $t$ as a root. But I don't know how to prove that it's irreducible over $F(t^n)$. I tried to consider that $t$ is trascendental to deduce some algebraic relations over $F$ on $t$. But it did not work.
75042
This is from Qing Liu's Algebraic Geometry and Arithmetic Curves (page 89). When he proves (c), he first prove the result when $K/k$ is finite, and then says that the general case immediately follows from the finite case and lemma 2.6 (posted below). I have no clue of how the lemma can be use to reduce the general case to the finite case.   Moreover, there is a remark pictured below that also uses the lemma 2.6 and once again I'm confused about how to use it. How can I complete these details?
24506
Consider the polynomial ring $F\left[x\right]$ over a field $F$. Let $d$ and $n$ be two nonnegative integers. Prove：$x^d-1 \mid x^n-1$ iff $d \mid n$.  my tries:  necessity, Let $n=d t+r$, $0\le r&lt;d$ since $x^d-1 \mid x^n-1$, so, $x^n-1=\left(x^d-1\right)\left(x^{\text{dt}+r-d}+\dots+1\right)$... so,,, to prove $r=0$? I don't know, and I can't go on. How to do it.
63744
Prove that:  $$a^{2^n} \equiv 1\pmod{2^{n+2}}$$  I have done the base case, and then have an assumption that:  $$a^{2^k} \equiv 1 \pmod {2^{k+2}}$$  I wish to prove the last step, that:  $$a^{2^{k+1}} \equiv 1 \pmod{2^{k+3}}$$  How would I do this?
113764
I'm stuck at the following question: let $d$ divide $2^{2^n} +1 $. Show that $d \equiv 1 \pmod{2^{n+1}}$.  All I've managed to do so far was to notice that $2^{n+1} \mid 2^{2^n}$ for every natural $n$, so if $dl = 2^{2^n} + 1$, we have $dl \equiv 1 \pmod{2^{n+1}}$ - which merely states that $d$ is invertible, which is trivial since it is odd. This is homework, are there any hints?
98874
Let $f\in L^1_\mathbb C(\mathbb R^n)$. I once read, in one of my old exam, that if $\hat{f}(\mathbb R^n)\subset\mathbb R_+$, then $\hat{f}\in L^1(\mathbb R^n)$. As far as I remember, the professor gave an identity using mollifiers but I'm not 100% sure. Any hints on how to prove that ?  Thanks!
76046
Suppose $k\in\mathbb N$ and suppose $2^k+1=u^2$ and $2^k-1=v^2$ for some $u,v\in\mathbb N$. I have to prove that the first case implies $k=3$ and the second case implies $k=1$. Also, for the second case it is hinted to use the fact (lemma) that if $n$ is odd, then $n^2\equiv1 \pmod 8$, but I suppose it could be used for the first one as well. My progress so far:     First question:      Suppose $u$ is even, then $u=2t$ for some $t\in\mathbb N$, thus $u^2=4t^2=2^k+1$. Notice that two divides the left hand side of the equation, therefore it also divides the right hand side. Since it divides $2^k$ for any $k\ge1$, it must also divide $1$, contradiction. So $u$ is odd and therefore $u^2\equiv1\pmod 8$. Thus there exists an $m\in\mathbb Z$ such that $u^2=8m+1=2^k +1$. Thus $2^k=8m$, therefore $k\ge3$.   I am stuck at this point, however. I suppose setting $k&gt;3$ and finding a contradiction would be a possible method, but I do not know what $k&gt;3$ actually has to offer to solve this problem. No other method I think of gets me anywhere, either. I guess I could try to show that $2^k+1$ cannot be a square for $k&gt;3$. Perhaps I could use the fact that $u^2-2^k=1$ or that $\gcd(u^2,2^k)=1$.     Second question:      The same reasoning can be applied to show that $v$ has to be even. Therefore, according to the lemma proposed above, $2^k-1=v^2\equiv1\pmod 8$. This implies $2^k\equiv2\pmod 8$. For any $k\ge3$ there exists some $t\in \mathbb N$ such that $2^k=8t$, therefore $2^k\equiv0\pmod 8$, which leads to a contradiction. Checking for $k=1,2$ leads to the conclusion that $k=1$ is the only possibility.   I think I have actually figured out this proof while writing down the progress I already have made, but for the sake of completeness and to check it's correctness I will leave it here.