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

It seems to me that since elementary algebra can be done almost mechanically when one has enough practice, it has been devaluated in the last decades. It's the same as with elementary arithmetic.

@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

@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.