More VW handout

by Wolstenholme, Nov 27, 2014, 4:13 AM

$1.2.a)$ OK so the key here is going to be the identity $\left(1 + x^{2^n}\right)\left(x^2 + x^4 + \dots + x^{2^n}\right) = x^2 + x^4 + \dots + x^{2^{n + 1}}.$

We proceed with induction on $n.$ The base case of $n = 1$ is trivial since we can let $p(x) = x^2$ and $q(x) = x.$

Now note that if $p, q \in \mathbb{Z}[x]$ such that $p(x)^2 + q(x)^2 = x^2 + x^4 + \dots + x^{2^n}$ we have $\left(p(x) - x^{2^n}q(x)\right)^2 + \left(x^{2^n}p(x) + q(x)\right)^2 = \left(1 + x^{2^n}\right)\left(p(x)^2 + q(x)^2\right) =$ $x^2 + x^4 + \dots + x^{2^{n + 1}}.$

Now since for any $2^n + 1$ -th root of unity $\omega$ we have $\omega^2 + \omega^4 + \dots + \omega^{2^{n + 1}} = -1$ we are done by induction.

To find this I basically just played with $n = 1$ and $n = 2$ and $n = 3$ and it took about 15 minutes.

$1.2.b)$ Ugh I solved this on the TSTST and I really don't want to do it again since the solution I had was awful. Basically by letting the coefficients of $D$ and $B$ be variables where $D(x)M(x) - B(x)C(x) = A(x)$ and expanding the LHS and setting the coefficient of every term with degree greater than $\frac{d}{2}$ to $0$ you get a system with more variables than equations so there exists a nontrivial solution.

$1.3)$ WLOG assume $\text{deg}(p(x)) \ge \text{deg}(q(x)).$ Assume for the sake of contradiction that $p(x) \ne q(x).$ Let $n = \text{deg}(p(x))$ and let $r$ be the number of distinct roots of $p(x)$ and let $s$ be the number of distinct roots of $p(x) + 1.$

First note that $p(x)$ and $p(x) + 1$ share no roots in common. Therefore $p(x) - q(x)$ has every distinct root of $p(x)$ and of $p(x) + 1$ as a root. This implies that $r + s \le \text{deg}(p(x) - q(x)) \le n.$

Now note that $p'(x)$ shares at least $n - r$ roots with $p(x)$ and at least $n - s$ roots with $p(x) + 1.$ Therefore $2n - r - s \le \text{deg}(p'(x) - q'(x)) \le n - 1$ which implies that $r + s \ge n + 1,$ contradiction.

This post has been edited 1 time. Last edited by Wolstenholme, Nov 27, 2014, 4:14 AM

Comment

0 Comments

Submit

glad to have found a fellow chipotle lover <3
by nukelauncher, Aug 13, 2020, 6:40 AM
the random chinese tst problem is the only thing I read, but I'll assume your blog is nice and give you a shout even though you probably never use aops anymoer
by fukano_2, Jun 14, 2020, 6:24 AM
wolstenholme - op
by AopsUser101, Jan 29, 2020, 8:27 PM
this blog is so hot
by mathleticguyyy, Jun 5, 2019, 8:26 PM
Hi. Nice Blog!
by User360702, Jan 10, 2019, 6:03 PM
helloooooo
by songssari, Jun 12, 2016, 8:21 AM
shouts make blogs happier
by briantix, Mar 18, 2016, 9:57 PM
You were just featured on AoPS's facebook page.
by mishka1980, Sep 12, 2015, 10:33 PM
This is late, but where is the ARML results post?
by donot, Aug 31, 2015, 11:07 PM
"I am Sam"
"Sam I am"
by mathwizard888, Aug 12, 2015, 9:13 PM
HW $\textcolor{white}{}$
by Eugenis, Apr 20, 2015, 10:10 PM
Uh-oh ARML practice is Thursday... I should start the homework.
by nosaj, Apr 20, 2015, 12:34 AM
Yes I am Sam, and Chebyshev polynomials aren't trivial, although they do make some problems trivial
by Wolstenholme, Apr 15, 2015, 10:00 PM
How are Chebyshev Polynomials trivial?
by nosaj, Apr 13, 2015, 4:10 AM
Are you Sam?
by Eugenis, Apr 4, 2015, 2:05 AM
@Brian: yes, yes I did #whoneedsalgskillz?

@gauss1181; hey!
by Wolstenholme, Mar 1, 2015, 11:25 PM
hello!!!
by gauss1181, Nov 27, 2014, 12:19 AM
Hi Wolstenholme did you actually use calc on that tstst problem
by briantix, Aug 2, 2014, 12:25 AM

18 shouts

Wolstenholme's blog

More VW handout

by Wolstenholme, Nov 27, 2014, 4:13 AM

0 Comments