let's start with the denominator: how did you get $(x-2)^2$? is that not $x^2-2x+4$? You instead start with $x^4-2x^2+1$ in the denominator

which is $(x^2)^2-2(x^2)+1=((x^2)-1)^2$

@PeterTamaroff I have 610 answers now.

I broke 666 recently

I have x^4 -2x^2 + 1 in the denominator which is (x-1)^2 isnt it?

@robjohn Cool!

I'll get to 300 in a while I guess.

Got 296.

@Jordan Close, it's $(x^{\color{Red}2}-1)^2$.

@anon That is a goal I am waiting for >8(

yes that is what I meant

so that is why I had that

but that leads me to $\int \frac{x^2}{x^2-1}$ which I am not sure is possible

Why not? Write the top as $$ (x^2-1)+1$$ and distribute,

You should end up with $$\int\sqrt{\frac{(x^2+1)^2}{(x^2-1)^2}}=\int \frac{x^2+1}{x^2-1}dx $$

why +1?

yeah that is what I got

@anon lol

and I split the two up I already solved the 1/x^2+ 1

@PeterTamaroff Bob Saget

@anon Eugene says IWBTG is a masochistic game. Do you concur?

I gonna win the game?

I think you mean IWBTG 8)

@anon Yes, sure,

Yes. But like I said, if you're gonna play it, go up from the first screen first.

well how do I solve the x^2 part?

I think the game was totally worth it though. At least halfway.

@Jordan Not sure what you mean. From where you should stand, you can write $$\frac{x^2+1}{x^2-1}=\frac{(x^2-1)+2}{x^2-1}=1+\frac{2}{x^2-1}=1+\frac{1}{x-1}-\frac{1}{x+1}.$$

I am bad with latex today.

why not $\frac{x^2}{x^2-1} + \frac{1}{x^2 - 1}$

Because what are you going to do with $\frac{x^2}{x^2-1}$?

I do not know

that is where I am stuck

it seems easier that that algebraic manipulation though

the technique of "adding and subtracting" in the numerator in order to do a cancellation and subsequent simplification is pretty useful

it isnt something I am use to though, I would always do the most simple and basic step frist

which to me would always be to split the fraction into two fractions

@Jordan What anon is doing is easier.

But it isn't something I would have been able to do on my own

I did split it into two fractions. But instead of the most simple and basic way, I took the most fundamental way, so that the numerator becomes non-polynomial. In order to make the numerator non-polynomial we need to get the $x^2$ out, and in order to do that we need to cancel a factor of $x^2-1$ with the denominator, hence we write $x^2+1=(x^2-1)+2$.

actually that is wrong isn't it? They differ by -4x as well

-4x^2

what are you talking about?

oh nevermind

I thought you did something different

why is it 1/x-1 - 1/x+1?

that's partial fraction decomposition on $2/(x^2-1)$

oh I am not good with that

Well, I have two quadratic cases memorized: $$\frac{1}{x(x+a)}=\frac{1}{a}\left(\frac{1}{x}-\frac{1}{x+a}\right)\qquad \frac{1}{(x-a)(x+a)}=\frac{1}{2a}\left(\frac{1}{x-a}-\frac{1}{x+a}\right).$$ One can manipulate these to decompose arbitrary reciprocals of quadratics.

It's the right formula that's of use here.

I guess the solution is very algebraic.

linear algebra!

@anon Yep!!

Ugh, my rep is 8,699

I should accept an answer, then downvote-

@anon Hm. I'm somehow getting the $ad-bc$ hint, but I'm not sure where I should end up in.

I have $$cm = a \pm y$$

I guess I can get the same for $n$

Errata,$$na = mc \pm y$$

algebra is really hard, it was never really taught in school

Try it with $(m,n)=1$ first. Try to exhibit a pair $u,v$ such that $um+vn=1$ by inverting the matrix $\begin{pmatrix}a&b\\c&d\end{pmatrix}$.

@anon OK. But I have a doubt on this:

$(m,n)=(x,y)$ implies $mX+nY=xX'+yY'$ or $mX+nY=xX+yY$?

@Jordan ((4x²)/(x⁴-2x²+1))+1=((x²+1)/(x²-1))²

the former

(bleh)

OK.

Should I be considering $\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$

?

mmmhmm

That is $$\begin{pmatrix}ax+by\\cx+dy\end{pmatrix}$$

now invert the system: write (x,y) as an integer matrix times (m,n)

@Jordan In that case I imagine it is.

@anon What do you call an integer matrix? All the $a_{ij} \in \Bbb N$?

$\in\Bbb Z$

@anon He right.

So $$\begin{pmatrix}a&b\\c&d\end{pmatrix}^{-1}=\pm \begin{pmatrix}d&-b\\-c&a\end{pmatrix}$$

the \pm shouldn't be there, this isn't SL_2(Z)

$SL_2(Z)$?

But we have that $ad-bc=\pm 1$

oh, right

@anon: I think that my argument is pretty minimal to show the case for $\mathbb{Z}$. I have to look closer at your argument for $\mathbb{Z}[x]$ though.

@AméricoTavares True. The art of teaching is very poor these days.

@anon: Norbert's is also essentially a polynomial argument.

Given $$\begin{pmatrix}m\\n\end{pmatrix} = \begin{pmatrix}a&b\\c&d\end{pmatrix} \begin{pmatrix}x\\y\end{pmatrix} , \tag{1}$$ we have $$\begin{pmatrix}x\\y\end{pmatrix}=\pm\begin{pmatrix}d&-b\\-c&a\end{pmatrix}\begin{pmatrix}m\\n\end{pmatrix}.\tag{2}$$ Now $(1)\implies(x,y)|m,n$ and conversely $(2)\implies (m,n)|x,y$. Hence $(x,y)|(m,n)$ and $(m,n)|(x,y)$.

@PeterTamaroff I think that the average student needs to be smarter than before to perform reasonably well when (s)he has to apply routine algebra, without having that background.

@AméricoTavares True.

@anon I may have missed, is $ad-bc=\pm1$?

yes

@PeterTamaroff And that is particularly true, e.g. in engineering, where a lot of trigonometry and algebra is applied.

@anon nm I see it

@robjohn Yes.

@anon I think that solution is really really nice. Kudos!

@anon: should I append a short justification that if $(a,c)=1$ and $c|ab$ then $c|b$?

append to what? the uniqueness proof of square/squarefree decomposition?

oh, in your answer

whatever you feel like mate

Going through school all I learned up to was geometry until I got to college, I never really learned algebra, trig or anything except really basic math

@Jordan My experience was just the opposite.

private or public schools?

@anon I just reorganized a bit. I don't think it's needed now.

@Jordan Not that I think that your reputation change is important to you, but you are now at the top 6% this month.

What does that mean?

@Jordan I think that just mean you are making progresses and asking many questions.

I see that now, I asked too many questions :P

@Jordan But if you have time enough, you should try and repeat the same integrals.

I actually have been going through them, I keep the question open in another tab, I haven't turned off my computer in 6 days because I have been slowly working through and adding more to review

@PaulSlevin oh don't feel that way! it really was alright.

@Jordan I understand. I think there is no hurry to accept answers.

@Eugene Hey Eugene! I've been making some progress with anon here, on Ch1 of Apostol.

I got 11/30 excersices.

@anon @PeterTamaroff please read henning's brilliant answer!

@PeterTamaroff Which book? Mathematical Analysis?

analytic number theory

@PeterTamaroff cool

@anon Do you know if there is a book with answers and some solutions to his analytic number theory's book?

@AméricoTavares i do not think there is one

no idea

i just love henning's response.

the last sentence is a little harsh though.

I read that there was such a book or some kind of copies for his two Calculus books.

@AméricoTavares Nope, Introduction to Analytic Number Theory

@AméricoTavares I have it.

Want it? (.pdf)

@Eugene LOL!

That is an awesome answer.

he deserves my upvote.

@AméricoTavares Wait. His Calculus books already contain the solutions.

@PeterTamaroff I think they were copies of some solutions given to his students.

@AméricoTavares Oh, OK. Anyways, I have the "solucionario" to his Analysis Book.

I am sorry, I meant his Mathematical Analysis book

@AméricoTavares I have it, do you need it?

@PeterTamaroff I would appreciate it very much

@PeterTamaroff I have so commented :-)

@robjohn LOL, upvoted that comment.

@AméricoTavares Mail?

Or maybe, wait.

Give me a sec.

Yes it can be. Look at my profile page.

@Eugene Isn't a little stupid to ask to prove $H_n=\sum_{k=1}^n \frac{1}{k}$ is never an integer?

How is that a stupid question? (also: n>1)

@PeterTamaroff where?

@Eugene Apostol.

@PeterTamaroff how is that a stupid question?

heh

@anon Well, $\gcd(1,2,\dots,n)=1$

man i thought $k$ was fixed for a moment there.

@Eugene LOL

@PeterTamaroff anyway, it's not super easy.

http://math.stackexchange.com/questions/2746/is-there-an-elementary-proof-that-sum-limits-k-1n-frac1k-is-never-an-int

@anon i'm confused, aren't the $p$-adic like that by definition?

@PeterTamaroff I have a proof here problemasteoremas.wordpress.com/2010/11/24/…

@Eugene Are you referencing my comment on the main?

@anon no i'm talking about the question

yeah, but it's possible someone defined p-adics differently to the OP

i see

@PeterTamaroff how do you intend to use this, exactly?

there's definitely no need for a $p$-adic number theory tag!

how is that not p-adic number theory?

this is just asking for the definition of $p$-adics. they're not an exclusively number theoretic concept.

why did you leave the number-theory tag?

nuts

thanks

i'm not sure i agree with topological groups either... but i don't know what to tag it as.

@AméricoTavares Cool. It is funny how knowing spanish one can almost understand portuguese.

@anon suggestions?

I'm still convinced p-adics are by default in the intersection of number theory and topological groups.

ok then

And anywhere else they appear they are simply being outsourced :)

how do you rollback?

go into the edit history and click rollback on the desired draft version

@anon i guess i don't have a rollback button then

i learn about $p$-adics in a geometry class.

geometry class?

what kind of geometry?

my friend learnt it in a galois theory class.

galois theory is sort of a building block in number theory on my view

ok then

anyway I put a group-theory tag in there, as the OP originally had one

instead of topological-groups

@anon oh ok.

I now clean my hands of this business.

lol

@Eugene That reminded me of youtube.com/watch?v=Le5nB5Bi6K4

@PeterTamaroff what did?

@Eugene "Ok, then."

ok then.

"Oh, thank God."

@PeterTamaroff The same for me with Spanish.

@AméricoTavares I'll give you that file now.

Here it is.

@PeterTamaroff Many thanks!

@MarianoSuárezAlvarez ?

@Eugene Are you around?

yes why?

@Eugene $$(a,b)=1$$ and $ab=c^n$ prove $a=x^n$ and $b=y^n$ for some $y$ and $x$.

Doesn't that follow from the unique factorization?

nope

this one is tough i think

i remember this being tough

this is a FTA dependent proof though

I mean, if $a=\prod p_i^{a_i}$ and $b=\prod p_j^{b_j}$ then $c^n=\prod p_i^{a_i} p_j^{b_j}$

Now $p_j \neq p_i$ for any subindex.

@PeterTamaroff ah

they're coprime

then should be ok

Yes yes.

man i keep on missing details these past two days

i need some sleep

@Eugene Ha, get some then!

Cheers.

@PeterTamaroff i've slept 8 hours in 3 days

@Eugene Apostol says consider $d=(a,c)$.

@Eugene Why?

@PeterTamaroff just doing my own projects now

I replenished my sleep this weekend. I slept very little in the week.

@PeterTamaroff good for you.

actually i've seen some of your answers

you're quite good at anal

@Eugene Hhahahahahahahha LOL

@PeterTamaroff =)

@Eugene I would star that, but I'd be weird.

it would be. i was hoping you wouldn't

@Eugene HOw would one exploit Apostol's hint on the last problem, namely considering $d=(a,c)$?

If you want don't answer that, I have a proof already.

isn't that just considering their common primes?

@Eugene Yep. Basically what I'm writing then.

@PeterTamaroff think you also have to consider $(b,c)$ no?

@Eugene I'm not doing it that way.

I simply write this

@PeterTamaroff huh. ok then

$$ab = \prod {{p_j}^{{b_j}}{p_i}^{{a_i}}} = {c^n} = \prod {{p_m}^{n{c_m}}} $$

Since $p_i \neq p_j$

We need

Thus

i disagree that this is a proof

you're assuming that the primes in $c^n$ are the same as those in $a$ and $b$.

you don't know that though.

and isn't it obvious is not an answer.

@Eugene But it follows from the unique factorization theorem.

@PeterTamaroff see here's the thing

$a$ has a unique factorization

$b$ has a unique factorization

and $c^n$ has a unique factorization

But $c$ is an integer.

but you need to show the primes are the same

Thus $c$ has a unique factorization.

that's why you're asked to consider the gcd

But doesn't $c$ have a unique factorization?

yes

but who says the primes are the same?

@PeterTamaroff For uniqueness of squarefree decomposition see here.

hi @BillDubuque!

$ab = c^n$ and every prime of $a$ is not in $b$, and $c$ has a unique factorization, plus $p_i ^n$ is never a prime. I don't see what is the problem.

@PeterTamaroff you need to prove the uniqueness. you're assuming it.

look at bill's link

The primes in $ab$ have to be the same as those in $c^n$, else we couldn't show equality.

@PeterTamaroff you have to show it man

hahahha

welcome to proofs!

@Eugene I proved what Bill writes already!

duh is not an answer

I proved that $n=a^2 b$ for $a$ $b$ unique for $n \geq 1$.

@PeterTamaroff not to me you didn't

@Eugene Hahah 'twas with Sire Anon.

@PeterTamaroff fine then

@PeterTamaroff wait

bill if responding to your earlier message??

then i disagree then

anyway

use apostol hint

@BillDubuque Bill, hello.

@Eugene OK. I guess this is wrong then:

$$n = \prod {{p_i}^{{a_i}}} $$

It can be the case, for some $a_i$, that $a_i = 2 c_i$. Then write

@PeterTamaroff no you're earlier answer is right.

$$n = {\left( {\prod {{p_i}^{{c_i}}} } \right)^2}\prod {{p_j}^{{a_j}}} $$

i just want you to make a clear answer on this

@Eugene ?

let's do this.

hm

let $c = \prod p_i ^{c_i}$

Now, for those $a_j = 2 b_j+1$ write $$n = {\left( {\prod {{p_i}^{{c_i}}{p_j}^{{b_j}}} } \right)^2}\prod {{p_j}} $$

@Eugene OK.

since $gcd(a,b) = 1$, $p_i$ divides either $a$ or $b$.

@Eugene Yes.