Mathematics

12:01 AM

Hey all! Can somebody drop me a hint for this question?
I'm unsure of where to even begin.
"Show that if $x$ and $y$ are non-zero integers and $x^2 =y^3$ then $x=a^3$ and $y=b^2$ for some integers $a$ and $b$."

Pedro Tamaroff

@KhallilBenyattou Note that any prime factor of $y$ is a prime factor of $x$ and vice versa.

Now consider prime factorizations.

One can do this nicely like this. Fix a prime number $p$, and let $\nu_p(x)$ denote the unique integer $n$ such that $p^n\mid x$ but $p^{n+1}\nmid x$.

Khallil Benyattou

May I ask which notation you use for writing out prime factorisations, @Pedro?

Pedro Tamaroff

@KhallilBenyattou Any understandable notation works =)

One checks that $\nu_p(x^m)=m\nu_p(x)$ for any integer $m\geqslant 0$.

Applying $\nu_p$ to your equation gives $2\nu_p(x)=3\nu_p(y)$. Since $\gcd(2,3)=1$, this gives that $3\mid \nu_p(x)$ and $2\mid \nu_p(y)$.

Say $\nu_p(x)=3k, \nu_p(y)=2j$. Then one gets $2\nu_p(x)=6j\implies \nu_p(x)=3j$ so $x$ is a cube, and similarily $\nu_p(y)=2k$, so $y$ is a square.

Khallil Benyattou

I've never seen a method like that!

Pedro Tamaroff

Valuations are good!

Khallil Benyattou

12:13 AM

Is that what they are called?

Pedro Tamaroff

@KhallilBenyattou Yes. In this case, to each rational number $r$ we may assign a (possibly negative!) integer $n$ such that $r=p^n r'$, and $r'$ is a rational where $p$ "doesn't appear".

Basically, write you rational number as a product of integer powers of primes, and take the integer this prime is risen to.

In this case I considered the valuation on the integers only.

Khallil Benyattou

I don't quite understand the second part. Why must we check the property that $v_p \left( x^m \right) = mv_p (x)$ for any integer $m \geqslant 0$?

Pedro Tamaroff

@KhallilBenyattou Because it is the useful part of the argument.

Applying $\nu_p$ to $x^2=y^3$ gives $\nu_p(x^2)=\nu_p(y^3)$.

Then I want to get the powers down.

$\nu_p$ satisfies $\nu_p(xy)=\nu_p(x)+\nu_p(y)$, so you can prove the above by induction.

Khallil Benyattou

For the last part, if $v_p(x)$ denotes the unique integer $n$ s.t. $p^n \mid x$ but $p^{n+1} \nmid x$ and $v_p(y)$ denotes the unique integer $m$ s.t. $p^m \mid y$ but $p^{m+1} \nmid y$, then we can conclude that $p^{n+m} \mid xy$, so $v_p(xy)$ denotes the unique integer $n+m$ s.t. $p^{n+m} \mid xy$, right @Pedro?

Pedro Tamaroff

@KhallilBenyattou Well, you need to show that $p^{n+m+1}\nmid xy$.

Hint If $n+m+1\mid xy$, then either $p$ has $n+1$ factors in $x$ or $m+1$ factors in $y$. Why?

(This is the pigeonhole principle!)

Khallil Benyattou

12:28 AM

May I ask why it's $-1$ in the exponent?

Pedro Tamaroff

That was a typo.

Khallil Benyattou

Also, do you mean $p^{n+m+1}$ instead of $n+m+1$ in the second line, @Pedro?

Pedro Tamaroff

@KhallilBenyattou Yes.

Ted Shifrin

@Pedro: You wanted to discuss thingies?

Pedro Tamaroff

@TedShifrin Yes.

Khallil Benyattou

12:34 AM

Is that because we've already established that $p^n \mid x$ and $p^{n+1} \nmid x$, then it's left that $p^{m+1} \mid y$ if $p^{n+m+1}$ indeed divides $xy$ (and the other argument is the same by interchanging $m$ and $n$), @Pedro?

Pedro Tamaroff

@KhallilBenyattou We already know that $n,m$ are respectively the largest integers $k$ such that $p^k$ divides $x,y$ respectively.

We want to see that $n+m$ is the largest integer $k$ such that $p^k$ divides $xy$.

It is clear $p^{n+m}\mid xy$. We want to see $p^{n+m-1}\nmid xy$.

Now if $p^{n+m+1}$ divides $xy$, consider two boxes, one for $x$ and one for $y$.

Khallil Benyattou

I thought that's what I was getting at with the contradiction that $p^{m+1} \mid y$ because by original definition for $v_p(y)$, $p^{m+1} \nmid y$, @Pedro.

Pedro Tamaroff

@KhallilBenyattou That's the point yes.

Ted Shifrin

@Pedro, so? :)

Pedro Tamaroff

If we distribute the $n+m+1$ factors, either $x$ receives $n+1$ factors or $y$ receives $m+1$ factors: if $x$ receives at most $n$ and $y$ receives at most $m$, we would account only for $n+m<n+m+1$ factors, @KhallilBenyattou

@TedShifrin I wanted to discuss the last exercise in your final.

How to show there are infinitely many roots.

Ted Shifrin

12:39 AM

Yes, so I presumed.

You need to learn what order of an entire function is, and decide what order that function has.

Khallil Benyattou

Ah, I kinda see what you mean. Thank you, @Pedro! I'll try and write up a solution to it. ^_^

Ted Shifrin

have fun @Khallil

Pedro Tamaroff

@KhallilBenyattou No problem.

@TedShifrin I've just read what the order of an entire function is. So that function has order... $1/2$?

Wait.

Ted Shifrin

Yup (make sure you prove it) :)

Pedro Tamaroff

OK, good.

I am guessing functions with order $<1$ have infinitely many zeros?

Ted Shifrin

12:43 AM

Functions with non-integer order have infinitely many zeroes (indeed, take on every value infinitely often).

Pedro Tamaroff

Oh, cool.

Ted Shifrin

I'll let you figure that out?

Pedro Tamaroff

=O

Ted Shifrin

There is a real theorem here ...

Mathy Person

1:33 AM

does anyone here have any experience with basic graph theory (simple graphs only)?

r9m

@robjohn sensei can you check my answer to my own question here when you have time ? :-) I didn't receive any satisfactory answer of it for a long time !

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$Mathematics$ Mathematics