Room for Liad and Daniel Fischer

now i want to prove that if $v_2(a-b) = n$ then $v_2(a \ ^ {2} - b \ ^ {2}) = n+1$ . i showed that $2 \ ^{ n+1} $ divides $a \ ^{ 2} - b\ ^ {2}$ so $v_2(a\ ^ {2} - b \ ^ {2})\ge n+1$

Daniel Fischer

That's not unconditionally true. Write $a^2 - b^2 = (a-b)(a+b)$, and think about what $v_2(a-b) = n$ implies about $v_2(a+b)$.

Liad

what do you mean? i need to prove it

wait i forgot to say $n\ge 2$

Daniel Fischer

Ah ha!

Liad

:)

so i did showed that $2 \ ^{ n+1}$ divides this product

Daniel Fischer

Yes, but without further assumptions, it can be that $v_p(a^2 - b^2) > n+1$.

Liad

8:01 PM

$a+b = a-b +2b$

@DanielFischer that's for the equality?

Daniel Fischer

@Liad That's helpful for the whole problem of how $v_p(a-b)$ and $v_p(a^2 - b^2)$ are related.

Liad

this gives me that $2$ divides $a+b$

because a-b and 2b can be divided by 2

Daniel Fischer

Yes. And under what conditions is it divisible by $4$, by $8$, …?

Liad

only if b does

Daniel Fischer

Not quite. Since we're assuming $a \equiv b \pmod{4}$, $a+b$ is divisible by $4$ if $b$ is even. It can be divisible by higher powers of $2$ if - what?

Liad

8:11 PM

if a-b does and b does , doesnt it?

Daniel Fischer

Consider $a = 14$ and $b = 2$.

Liad

you are asking about a+b right?

Daniel Fischer

Yes.

Liad

$b = a - l 2 \ ^{n}$

so $a+b = 2a - l 2 \ ^ {n}$

so the sum is divisible by higher powers of 2 if a does

right?

Daniel Fischer

Both play a part in it.

Liad

8:21 PM

Proof by contradiction wont work?

i tried assuming $2 \ ^ {n+2}$ divides the product, couldn't continue..

Daniel Fischer

And unless you impose a certain restriction to $a$ and $b$, $v_2(a+b)$ can be as large as you want when $v_2(a-b) = n \geqslant 2$.

Liad

well i got that $2$ divides $(a-2 \ ^ {n-1}l)l$

(if i assume $2 \ ^ {n+2} $ divides the product)

so that means that $2$ divides $l$ or 2 divides $a-2 \ ^{ n-1}l$

but $2$ cant divide $l$

because $v_2(a-b) = n$

Daniel Fischer

You're thinking too complicated. Look at the example above, $a = 14, b = 2$, and see what happens there.

Liad

$v_2(a-b) = 2$

$v_2(a+b) =4$

Daniel Fischer

$a + b = 16 = 2^4$

Liad

8:30 PM

^^

i can blame the beer i just drank

for that mistake

Daniel Fischer

Now take for example $a = 252$ and $b = 4$.

Liad

$v_2(a+b) = 8$

$v_2(a-b) = 3$

Daniel Fischer

Right. Do you see that you can have $v_2(a+b)$ arbitrarily large?

Liad

yes i can see that

Daniel Fischer

Okay. So what can prevent that?

Liad

8:33 PM

Hmm

maybe to pick a,b that are close to each other?

Daniel Fischer

Not really. $130$ and $126$ are as close together as can be for $v_2(a-b) = 2$.

Look at $a+b = (a-b) + 2b$. What do you know about $v_p(x+y)$ in terms of $v_p(x)$ and $v_p(y)$?

Liad

why would we want to prevent $v_2(a+b)$ to be large?

by p you mean 2 or arbitrary p?

Daniel Fischer

@Liad Arbitrary $p$. Here we're interested in $p = 2$, but the principle is general.

@Liad Isn't the aim to have $v_2(a^2 - b^2) = v_2(a-b) + 1$ if $v_2(a-b) \geqslant 2$ - and some further condition?

Liad

its the aim (no further condition)

Daniel Fischer

And $v_2(a^2 - b^2) = v_2(a-b) + v_2(a+b)$.

Liad

8:39 PM

the next question is to show it for arbirary $p $ and the condition is $v_p(a-b)\ge1$

Daniel Fischer

@Liad Well, then you've seen counterexamples. Somebody forgot the additional condition needed.

Liad

@DanielFischer where is this come from? its not in the exercise

Daniel Fischer

@Liad Generally, $v_p(x\cdot y) = v_p(x) + v_p(y)$.

(follows from unique factorisation)

Liad

so they forgot to say that $v_2(a+b) = 1$ !?

Daniel Fischer

Well, indirectly. Stating it in that form would make the exercise rather pointless.

Liad

8:43 PM

didnt you say that without this condition this exercise is not true?

$a=14 , b=2$

$v_2(a-b) = 2$

$v_2(a \ ^ {2} - b \ ^ {2}) = v_2(192) = 6$

so they are wrong

Daniel Fischer

Considering the next exercise - which also requires the analogous condition - is there maybe a general restriction that $a,b$ aren't divisible by $p$?

Liad

Ahhhhhhhhhhhhhhhh

i did not read that

yes there is that condition.... it was stated before the exercise

didnt read it :/

sorry..

Daniel Fischer

Okay. So we have two odd numbers congruent modulo $4$. Deduce that their sum is not divisible by $4$.

Liad

yea now it works.

cant believe i missed that condition

shouldn't drink and do math ^^

Daniel Fischer

Old saying, "Don't drink and derive".

Liad

8:52 PM

hehe nice

Daniel Fischer

For general $p$, it's probably $v_p(a^p - b^p) = v_p(a-b) + 1$?

Liad

yes and this time $v_p(a-b) \ge 1$

$p \ne 2$

Daniel Fischer

I suggest you don't do that by factoring $a^p - b^p = (a-b)(\dotsc)$, rather write $a = b + k\cdot p^n$ where $p \nmid k$, and use the binomial theorem.

Liad

the $b \ ^ {p} $ will cancel that's for sure

^^

i will work it out tomorrow. hopefully because i have this c++ exercise that i must start too

thanks again

Good night

Daniel Fischer

Laila tovah

Liad

9:03 PM

tov

are you from israel?