HSC 2016 MX2 Marathon ADVANCED (archive) (5 Viewers)

Zen2613 · Oct 28, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Drsoccerball said:
If you noticed I commented on the unedited version of his post. The unedited one had $A^2 =2 (integer)$ Which wouldnt allow A to be a multiple of 2 for all real values of the RHS.

I think the other guy's proof is valid. You have A^2 = 2X, where A and X are both integers. Now consider if A is odd, i.e A = 2n+1 then A^2 = 4n^2 + 4n + 1 = 2(2n^2 + 2n) + 2 = 2m + 1 (n, m are integers). Therefore if A is odd, the A^2 is odd which is a contradiction, so A can't odd. If A is even then A = 2n, then A^2 = 2(2n^2) = 2m i.e A^2 is even if A is even. This works and therefore A must be even QED. No ? The only other possibility is if A is not an integer, but it is stated in the first place that it is, so the proof holds.

InteGrand · Oct 28, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Drsoccerball said:
If you noticed I commented on the unedited version of his post. The unedited one had $A^2 =2 (integer)$ Which wouldnt allow A to be a multiple of 2 for all real values of the RHS.

$That's fine though. If $n$ is an integer and $n^2$ is even (that is, $n^2 = 2k$ for some integer $k$), then $n$ must be even.$

InteGrand · Oct 28, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Drsoccerball said:
You dont have to assume most of those results as they are given to be positive.

Another reason you don't need to assume some of them is that negative integers can be even and odd too, so you only need to know these three: odd + odd = even, odd + even = odd, even + even = even. (Or even simpler, if the two numbers being added have the same parity, the sum is even, and if the two numbers being added have different parity, then the sum is odd.)

lita1000 · Oct 28, 2015

Re: HSC 2016 4U Marathon - Advanced Level

InteGrand said:
$That's fine though. If $n$ is an integer and $n^2$ is even (that is, $n^2 = 2k$ for some integer $k$), then $n$ must be even.$

Exactly my point. Drongoski was right all along

lita1000 · Oct 28, 2015

Re: HSC 2016 4U Marathon - Advanced Level

InteGrand said:
Another reason you don't need to assume some of them is that negative integers can be even and odd too, so you only need to know these three: odd + odd = even, odd + even = odd, even + even = even. (Or even simpler, if the two numbers being added have the same parity, the sum is even, and if the two numbers being added have different parity, then the sum is odd.)

Can you or sy or someone post a difficult problem?

InteGrand · Oct 28, 2015

Re: HSC 2016 4U Marathon - Advanced Level

lita1000 said:
Can you or sy or someone post a difficult problem?

You can just do Sy123's one from the first page, since no-one's answered it yet:

Sy123 said:
$\\ $Show that the only pairs of positive integers$ \ x , y \ $so that$ \\ x^y = y^x \ $and$ \ x \neq y \\ $are$ \ x = 2, y = 4 \ $and$ \ x = 4, y = 2$

lita1000 · Oct 28, 2015

Re: HSC 2016 4U Marathon - Advanced Level

InteGrand said:
You can just do Sy123's one from the first page, since no-one's answered it yet:

Notice that the derivative of lnx /x is (1-lnx)/x^2, which <0 if x>e, so f(x)= lnx /x is a strictly decreasing function for x>e. Thus if y>x>e, there is no positive integer solutions for x^y=y^x (since this rearranges to form lnx/x=lny/y). So we only have to test for integers between 0 and e.
if x= 1 then trivially y=1 but since x=/=y this doesn't count. If x= 2, the only positive integer solution for y=4. By symmetry, (x,y)

4,2) is another solution.

Sy123 · Oct 28, 2015

Re: HSC 2016 4U Marathon - Advanced Level

lita1000 said:
Notice that the derivative of lnx /x is (1-lnx)/x^2, which <0 if x>e, so f(x)= lnx /x is a decreasing function for x>e. Thus if y>x>e, there is no positive integer solutions for x^y=y^x. So we only have to test for integers in 0<x<e, if x= 1 then trivially y=1 but since x=/=y this doesn't count. If x= 2, the only positive integer solution for y=4. By symmetry, (x,y)4,2) is another solution.

Well done

lita1000 · Oct 28, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Sy123 said:
Well done

Do you have a difficult problem to share with us?

Sy123 · Oct 28, 2015

Re: HSC 2016 4U Marathon - Advanced Level

lita1000 said:
Do you have a difficult problem to share with us?

All topics?

Some ones that I can think of on the top of head that don't require any 4U knowledge:

$\\ $Prove that if and only if the sum of the digits of a number are divisible by 3, then the number is divisible by 3$

InteGrand · Oct 28, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Sy123 said:
All topics?

Some ones that I can think of on the top of head that don't require any 4U knowledge:

$\\ $Prove that if and only if the sum of the digits of a number are divisible by 3, then the number is divisible by 3$

$Let the number $N$ be $N = a_n \times 10^n + a_{n-1}\times 10^{n-1} + \cdots + a_1 \times 10 + a_0$, so in digit form, $N$ is $a_n a_{n-1}...a_1 a_0$. Clearly, for any natural number $k$, we have $10^k \equiv 1 \pmod 3$, as $10\equiv 1 \pmod 3$. Therefore, using modular arithmetic rules, we have$

$\begin{align*}N=a_n \times 10^n + a_{n-1}\times 10^{n-1} + \cdots + a_1 \times 10 + a_0 &\equiv a_n \cdot 1 + a_{n-1}\cdot 1 + \cdots + a_1 \cdot 1 + a_0 \cdot 1 \pmod 3 \\ \Rightarrow N &\equiv a_n + a_{n-1}+\cdots + a_1 + a_0 \pmod 3 . \end{align*}$

$It follows that $N\equiv 0 \pmod 3$ if and only if $a_n + a_{n-1}+\cdots + a_1 + a_0 \equiv 0 \pmod 3$, that is, $N$ is divisible by 3 if and only if its digit sum is divisible by 3.$

Sy123 · Oct 28, 2015

Re: HSC 2016 4U Marathon - Advanced Level

InteGrand said:
$Let the number $N$ be $N = a_n \times 10^n + a_{n-1}\times 10^{n-1} + \cdots + a_1 \times 10 + a_0$, so in digit form, $N$ is $a_n a_{n-1}...a_1 a_0$. Clearly, for any natural number $k$, we have $10^k \equiv 1 \pmod 3$, as $10\equiv 1 \pmod 3$. Therefore, using modular arithmetic rules, we have$

$\begin{align*}N=a_n \times 10^n + a_{n-1}\times 10^{n-1} + \cdots + a_1 \times 10 + a_0 &\equiv a_n \cdot 1 + a_{n-1}\cdot 1 + \cdots + a_1 \cdot 1 + a_0 \cdot 1 \pmod 3 \\ \Rightarrow N &\equiv a_n + a_{n-1}+\cdots + a_1 + a_0 \pmod 3 . \end{align*}$

$It follows that $N\equiv 0 \pmod 3$ if and only if $a_n + a_{n-1}+\cdots + a_1 + a_0 \equiv 0 \pmod 3$, that is, $N$ is divisible by 3 if and only if its digit sum is divisible by 3.$

For people unfamiliar with modular arithmetic:

$\\ N = 10^n a_n + \dots + 10a_1 + a_0 \ $is a representation of any number with digits$ \ a_n, a_{n-1}, \dots , a_0$

$\\ $Let$\ k = a_n + a_{n-1} + \dots + a_0$

$\\ N-k = \sum_{i=1}^n (10^i - 1)a_i = \sum_{i=1}^n (1+10+\dots + 10^{i-1})(10-1)a_i = 3 \sum_{i=1}^n 3a_i (1+10 + \dots + 10^{i-1}) = 3M$

$\\ $Thus$ \ N - k \ $is divisible by$ 3$

$\\ $So, if$ \ N \ $is divisible by 3, then$ \ N - 3M = 3 (N/3 - M) = k \ $then$ \ k \ $is divisible by 3$$

$\\ $If conversely$ \ k \ $is divisible by 3, then$ \ N = 3(M + k/3) \ $then$ \ N \ $is divisible by 3$$

Sy123 · Oct 28, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Here is a good question (not too mechanical)

$\\ $Two polynomials$ \ P(x) \ $and$ \ Q(x) \ $have real co-efficients and are such that$ \ P(x) = Q(x) \ $for real values of$ \ x$

$\\ $Prove$ \ P(x) = Q(x) \ $for all complex values of$ \ x$

lita1000 · Oct 28, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Sy123 said:
Here is a good question (not too mechanical)

$\\ $Two polynomials$ \ P(x) \ $and$ \ Q(x) \ $have real co-efficients and are such that$ \ P(x) = Q(x) \ $for real values of$ \ x$

$\\ $Prove$ \ P(x) = Q(x) \ $for all complex values of$ \ x$

Consider P(x)-Q(x)= R(x), this has infinite roots, and the only polynomial with infinite roots is R(x)=0. Thus, R(z) is also =0 for all complex numbers z, and hence P(x)=Q(x) for all complex numbers as well

glittergal96 · Oct 28, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Related:

Find all polynomials p(z) such that p(cis(t))=p(cis(t+a)) for all t, where a is a given positive number.

You may need to consider cases.

Sy123 · Oct 29, 2015

Re: HSC 2016 4U Marathon - Advanced Level

glittergal96 said:
Related:

Find all polynomials p(z) such that p(cis(t))=p(cis(t+a)) for all t, where a is a given positive number.

You may need to consider cases.

$\\ p(z(cos \alpha + i \sin \alpha)) = p(z) \Rightarrow \ p(z(cos k\alpha + i \sin k\alpha)) = p(z) \ $for all integers$ \ k \geq 0$

$\\ \alpha \ $is either a rational multiple of$ \ 2\pi \ $or not, if it is not, then$ \ \cis k \alpha \ $is distinct for all values of$ \ k \ $(L1)$$

$\\ $Since$ \ p(1) = p(cis k \alpha) \ $for all values of$ \ k \ $then$ \ p(z) - p(1) \ $has infinitely many roots, meaning$ \ p(z) = C \ $for some constant$ \ C$

$\\ $Suppose on the other hand that$ \ \alpha = 2\pi\frac{p}{q} \ $then$ \\\\ p(t) = p\left(t \cdot cis \frac{2\pi p}{q} \right) = \dots = p \left(t \cdot cis \frac{2\pi (q-1)p}{q} \right)$

$\\ $Then$ \ p(z) - p(t) \ $has$ \ q \ $roots exactly$$

$\\ $Then$ \ p(z) - p(t) = A(z-t)\left(z-t\cdot cis \frac{2\pi}{q} \right) \dots \left(z-t\cdot cis \frac{2\pi}{q}(q-1) \right) = A(z^q - t^q)$

$\\ $Then$ \ p(z) = A z^q + B \ $for some constants$ \ A ,B$

$\\ $Thus, the final solution is$ \ p(z) = \begin{cases}c , \ \alpha \ $is not a rational multiple of$ \ 2\pi \\ a z^q + b , \ \alpha \ $is a rational multiple of$ \ 2\pi \ $with denominator$ \ q \end{cases}$

-----------

$\\ $(L1) proof as follows:$ \ $Suppose there exists$ \ m < n \ $so that$ \ cis m \alpha = cis n \alpha \ $with$ \ m,n \ $natural$ \\ $Then$ \ m\alpha + 2\pi k = n\alpha \ $for some integer$ \ k \ $as$ \ cis \ $is periodic in$ \ 2\pi \\ \\ $So,$ \ \alpha = 2\pi \frac{k}{m-n} \ $which is a rational multiple of$ \ 2\pi \\\\ $Contradiction, therefore$ \ $all distinct if not rational multiple$$

glittergal96 · Oct 29, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Sy123 said:
$\\ p(z(cos \alpha + i \sin \alpha)) = p(z) \Rightarrow \ p(z(cos k\alpha + i \sin k\alpha)) = p(z) \ $for all integers$ \ k \geq 0$

$\\ \alpha \ $is either a rational multiple of$ \ 2\pi \ $or not, if it is not, then$ \ \cis k \alpha \ $is distinct for all values of$ \ k \ $(L1)$$

$\\ $Since$ \ p(1) = p(cis k \alpha) \ $for all values of$ \ k \ $then$ \ p(z) - p(1) \ $has infinitely many roots, meaning$ \ p(z) = C \ $for some constant$ \ C$

$\\ $Suppose on the other hand that$ \ \alpha = 2\pi\frac{p}{q} \ $then$ \\\\ p(t) = p\left(t \cdot cis \frac{2\pi p}{q} \right) = \dots = p \left(t \cdot cis \frac{2\pi (q-1)p}{q} \right)$

$\\ $Then$ \ p(z) - p(t) \ $has$ \ q \ $roots exactly$$

$\\ $Then$ \ p(z) - p(t) = A(z-t)\left(z-t\cdot cis \frac{2\pi}{q} \right) \dots \left(z-t\cdot cis \frac{2\pi}{q}(q-1) \right) = A(z^q - t^q)$

$\\ $Then$ \ p(z) = A z^q + B \ $for some constants$ \ A ,B$

$\\ $Thus, the final solution is$ \ p(z) = \begin{cases}c , \ \alpha \ $is not a rational multiple of$ \ 2\pi \\ a z^q + b , \ \alpha \ $is a rational multiple of$ \ 2\pi \ $with denominator$ \ q \end{cases}$

-----------

$\\ $(L1) proof as follows:$ \ $Suppose there exists$ \ m < n \ $so that$ \ cis m \alpha = cis n \alpha \ $with$ \ m,n \ $natural$ \\ $Then$ \ m\alpha + 2\pi k = n\alpha \ $for some integer$ \ k \ $as$ \ cis \ $is periodic in$ \ 2\pi \\ \\ $So,$ \ \alpha = 2\pi \frac{k}{m-n} \ $which is a rational multiple of$ \ 2\pi \\\\ $Contradiction, therefore$ \ $all distinct if not rational multiple$$

Very close.

Why must p(z)-p(t) have exactly q roots in the rational multiple of pi case? Also are you saying this is true for all t? Or some particular choice of t?

I agree that if t is nonzero, then p(z)-p(t) must have at LEAST q roots.

Sy123 · Oct 29, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Revamping solution (and making some of the inferences more rigorous), had fun writing this

glittergal96 said:
Related:

Find all polynomials p(z) such that p(cis(t))=p(cis(t+a)) for all t, where a is a given positive number.

You may need to consider cases.

$\\ p(z(\cos \alpha + i \sin \alpha)) = p(z) \\\\ \Rightarrow \ p(z(\cos k\alpha + i \sin k\alpha)) = p(z) \ $for all integers$ \ k \geq 0$

$\\ \alpha \ $is either a rational multiple of$ \ 2\pi \ $or not, if it is not, then$ \ \cos k \alpha + i \sin k \alpha \ $is distinct for all values of$ \ k \ $(L1)$$

$\\ $Since$ \ p(1) = p(\cos k \alpha + i \sin k \alpha) \ $for all values of$ \ k \ $then$ \ p(z) - p(1) \ $has infinitely many roots, meaning$ \ p(z) = C \ $for some constant$ \ C$

$\\ $Suppose on the other hand that$ \ \alpha = 2\pi\frac{p}{q} \ $then$ \\\\ p(t) = p\left(t \left( \cos \frac{2\pi p}{q} i \sin \frac{2\pi p}{q} \right) \right) = \dots = p \left(t \left( \cos \frac{2\pi p (q-1)}{q} i \sin \frac{2\pi p (q-1)}{q} \right)\right)$

$\\ $Then$ \ p(z) - p(t) = A(z-t)\left(z - t \left( \cos \frac{2\pi p}{q} i \sin \frac{2\pi p}{q} \right) \right) \dots \left(z - t \left( \cos \frac{2\pi p (q-1)}{q} i \sin \frac{2\pi p (q-1)}{q} \right) \right) M(z)$

$\\ $Where$ \ M(z) \ $is a non-zero polynomial$$

$\\ (z-t)\left(z - t \left( \cos \frac{2\pi p}{q} i \sin \frac{2\pi p}{q} \right) \right) \dots \left(z - t \left( \cos \frac{2\pi p (q-1)}{q} i \sin \frac{2\pi p (q-1)}{q} \right) \right) \\\\ = z^q - t^q \ $(L2)$$

$\\ $Then$ \ p(z) - p(t) = A(z^q - t^q) M(z)$

$\\ $Let$ \ z = t +h \ $for some small$ \ h > 0$

$p(t+h) - p(t) = A((t+h)^q - t^q) M(t+h)$

$\\ \Rightarrow p(t+h) - p(t) = Ah( (t+h)^{q-1} + t(t+h)^{q-2} + \dots + t^{q-1}) M(t+h)$

$\\ \Rightarrow p'(t) = A \lim_{h \to 0}( (t+h)^{q-1} + t(t+h)^{q-2} + \dots + t^{q-1}) M(t+h) = A \cdot M(t) q t^{q-1} \ \ (*)$

$\\ $Let instead$ \ z = 0 \ $then$ \ p(t) = A\cdot M(t) t^q + B \ \ (**)$

$\\ $Differentiating$ \ (**) \ $we get$ \ p'(t) = A( M'(t) t^q + M(t) q t^{q-1})$

$\\ $Equating$ \ (**) \ $and$ \ (*) \ $we get$ \ M'(t) = 0 \Rightarrow M(t) = C$

$\\ $Therefore$ \ p(z) - p(t) = A(z^q - t^q)$

$\\ \Rightarrow \ p(z) = a z^q + b \ $for constants$ \ a,b$

$\\ $Thus, the final solution is$ \ p(z) = \begin{cases}c , \ \alpha \ $is not a rational multiple of$ \ 2\pi \\ a z^q + b , \ \alpha \ $is a rational multiple of$ \ 2\pi \ $with denominator$ \ q \end{cases}$

-----------

$\\ $(L1) proof as follows:$ \ $Suppose there exists$ \ m < n \ $so that$ \ cis m \alpha = cis n \alpha \ $with$ \ m,n \ $natural$ \\ $Then$ \ m\alpha + 2\pi k = n\alpha \ $for some integer$ \ k \ $as$ \ cis \ $is periodic in$ \ 2\pi \\ \\ $So,$ \ \alpha = 2\pi \frac{k}{m-n} \ $which is a rational multiple of$ \ 2\pi \\\\ $Contradiction, therefore$ \ $all distinct if not rational multiple$$

-----------

$\\ $(L2) proof as follows: It is sufficient to prove that$ \\\\ \prod_{i=0}^{q-1} \left(z - t\left(\cos \frac{2\pi i p}{q} + i\sin \frac{2 \pi i p}{q} \right) \right) = \prod_{i=0}^{q-1} \left(z - t\left(\cos \frac{2\pi i}{q} + i\sin \frac{2 \pi i }{q} \right) \right)$

$\\ $This can be done by showing that$ \\\\ $(1):$ \ \cos \frac{2\pi i p}{q} + i\sin \frac{2 \pi i p}{q} = \cos \frac{2\pi j}{q} + i\sin \frac{2 \pi j}{q} \ $for some integers$ \ i,j \\\\ $(2): From the above equation, for each different$ \ i \ $there is a unique$ \ j$

$\\ $(1) and (2) then imply that the set of: $ \\\\ \left(z - t\left(\cos \frac{2\pi i p}{q} + i\sin \frac{2\pi i p}{q} \right) \right) \ $is a permutation of$ \\\\\ \left(z - t\left(\cos \frac{2\pi i}{q} + i\sin \frac{2 \pi i }{q} \right) \right)$

$\\ $Meaning their products are equal$$

$\\ $Proof of (1) is that for each$ \ ip = r_i + k_iq \ $for some integers$ \ 0 \leq r_i \leq q-1 \ $and integer$ \ k_i$

$\frac{2\pi ip}{q} = \frac{2\pi r_i}{q} + 2\pi k_i$

$\\ $Since$ \ \cos \theta + i \sin \theta \ $is periodic in$ \ 2\pi \ $then$$

$\cos \frac{2\pi i p}{q} + i\sin \frac{2 \pi i p}{q} = \cos \frac{2\pi j}{q} + i\sin \frac{2 \pi j}{q} \ $where$ \ j = r_i$

$\\ $Proof of (2) is that for any two distinct integers$ \ m,n \ $then if$ \ mp = r_m + k_mq \ $and$ \ np = r_n + k_nq \ $then$ \ r_n \neq r_m \ $as it is the remainders that get mapped onto set of$ \ j$

$\\ $Suppose to the contrary that$ \ r_m = r_n \ $then$ \ \frac{(m-n)p}{q} = K \ $for some integer$ \ K$

$\\ $Given the definition of$ \ p,q \ $then$ \ p \ $and$ \ q \ $have no common factors, moreover, since$ \ 0 < m-n < q-1 \ $then$ \ (m-n) \ $is not divisible by$ \ q \ $then in order for$ \ \frac{(m-n)p}{q} = K \ $all the factors of$ \ q \ $must divide into either$ \ (m-n) \ $or$ \ p \ $but since it can't be$ \ p \ $then it must be$ \ (m-n)$

$\\ $But not all the factors of$ \ q \ $can divide into$ \ (m-n) \ $as that would mean$ \ (m-n) \ $is divisible by$ \ q \ $which is already shown to be impossible$$

$\\ $Therefore$ \ \frac{(m-n)p}{q} \neq K \ $thus$ \ r_m \neq r_n \ $thus$ \ (2) \ $is proven$$

$\\ $Thus the proposition in question is proven$$

-----------------

$\\ $NOTE: (*) means that at HSC level at least, this at most finds all$ \ \textbf{real} \ $polynomials with the aforementioned properties$$

lita1000 · Oct 29, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Sy123 said:
Revamping solution (and making some of the inferences more rigorous), had fun writing this

$\\ p(z(\cos \alpha + i \sin \alpha)) = p(z) \\\\ \Rightarrow \ p(z(\cos k\alpha + i \sin k\alpha)) = p(z) \ $for all integers$ \ k \geq 0$

$\\ \alpha \ $is either a rational multiple of$ \ 2\pi \ $or not, if it is not, then$ \ \cos k \alpha + i \sin k \alpha \ $is distinct for all values of$ \ k \ $(L1)$$

$\\ $Since$ \ p(1) = p(\cos k \alpha + i \sin k \alpha) \ $for all values of$ \ k \ $then$ \ p(z) - p(1) \ $has infinitely many roots, meaning$ \ p(z) = C \ $for some constant$ \ C$

$\\ $Suppose on the other hand that$ \ \alpha = 2\pi\frac{p}{q} \ $then$ \\\\ p(t) = p\left(t \left( \cos \frac{2\pi p}{q} i \sin \frac{2\pi p}{q} \right) \right) = \dots = p \left(t \left( \cos \frac{2\pi p (q-1)}{q} i \sin \frac{2\pi p (q-1)}{q} \right)\right)$

$\\ $Then$ \ p(z) - p(t) = A(z-t)\left(z - t \left( \cos \frac{2\pi p}{q} i \sin \frac{2\pi p}{q} \right) \right) \dots \left(z - t \left( \cos \frac{2\pi p (q-1)}{q} i \sin \frac{2\pi p (q-1)}{q} \right) \right) M(z)$

$\\ $Where$ \ M(z) \ $is a non-zero polynomial$$

$\\ (z-t)\left(z - t \left( \cos \frac{2\pi p}{q} i \sin \frac{2\pi p}{q} \right) \right) \dots \left(z - t \left( \cos \frac{2\pi p (q-1)}{q} i \sin \frac{2\pi p (q-1)}{q} \right) \right) \\\\ = z^q - t^q \ $(L2)$$

$\\ $Then$ \ p(z) - p(t) = A(z^q - t^q) M(z)$

$\\ $Let$ \ z = t +h \ $for some small$ \ h > 0$

$p(t+h) - p(t) = [B]A((t+h)^q - t^q) M(t+h)$

$\\ \Rightarrow p(t+h) - p(t) = Ah( (t+h)^{q-1} + t(t+h)^{q-2} + \dots + t^{q-1}) M(t+h)[/B]$

$\\ \Rightarrow p'(t) [B]= A \lim_{h \to 0}( (t+h)^{q-1} + t(t+h)^{q-2} + \dots + t^{q-1}) M(t+h) = A \cdot M(t) q t^{q-1} \ \ (*)$ [/B]

$\\ $Let instead$ \ z = 0 \ $then$ \ p(t) = A\cdot M(t) t^q + B \ \ (**)$

$\\ $Differentiating$ \ (**) \ $we get$ \ p'(t) = A( M'(t) t^q + M(t) q t^{q-1})$

$\\ $Equating$ \ (**) \ $and$ \ (*) \ $we get$ \ M'(t) = 0 \Rightarrow M(t) = C$

$\\ $Therefore$ \ p(z) - p(t) = A(z^q - t^q)$

$\\ \Rightarrow \ p(z) = a z^q + b \ $for constants$ \ a,b$

$\\ $Thus, the final solution is$ \ p(z) = \begin{cases}c , \ \alpha \ $is not a rational multiple of$ \ 2\pi \\ a z^q + b , \ \alpha \ $is a rational multiple of$ \ 2\pi \ $with denominator$ \ q \end{cases}$

-----------

$\\ $(L1) proof as follows:$ \ $Suppose there exists$ \ m < n \ $so that$ \ cis m \alpha = cis n \alpha \ $with$ \ m,n \ $natural$ \\ $Then$ \ m\alpha + 2\pi k = n\alpha \ $for some integer$ \ k \ $as$ \ cis \ $is periodic in$ \ 2\pi \\ \\ $So,$ \ \alpha = 2\pi \frac{k}{m-n} \ $which is a rational multiple of$ \ 2\pi \\\\ $Contradiction, therefore$ \ $all distinct if not rational multiple$$

-----------

$\\ $(L2) proof as follows: It is sufficient to prove that$ \\\\ \prod_{i=0}^{q-1} \left(z - t\left(\cos \frac{2\pi i p}{q} + i\sin \frac{2 \pi i p}{q} \right) \right) = \prod_{i=0}^{q-1} \left(z - t\left(\cos \frac{2\pi i}{q} + i\sin \frac{2 \pi i }{q} \right) \right)$

$\\ $This can be done by showing that$ \\\\ $(1):$ \ \cos \frac{2\pi i p}{q} + i\sin \frac{2 \pi i p}{q} = \cos \frac{2\pi j}{q} + i\sin \frac{2 \pi j}{q} \ $for some integers$ \ i,j \\\\ $(2): From the above equation, for each different$ \ i \ $there is a unique$ \ j$

$\\ $(1) and (2) then imply that the set of: $ \\\\ \left(z - t\left(\cos \frac{2\pi i p}{q} + i\sin \frac{2\pi i p}{q} \right) \right) \ $is a permutation of$ \\\\\ \left(z - t\left(\cos \frac{2\pi i}{q} + i\sin \frac{2 \pi i }{q} \right) \right)$

$\\ $Meaning their products are equal$$

$\\ $Proof of (1) is that for each$ \ ip = r_i + k_iq \ $for some integers$ \ 0 \leq r_i \leq q-1 \ $and integer$ \ k_i$

$\frac{2\pi ip}{q} = \frac{2\pi r_i}{q} + 2\pi k_i$

$\\ $Since$ \ \cos \theta + i \sin \theta \ $is periodic in$ \ 2\pi \ $then$$

$\cos \frac{2\pi i p}{q} + i\sin \frac{2 \pi i p}{q} = \cos \frac{2\pi j}{q} + i\sin \frac{2 \pi j}{q} \ $where$ \ j = r_i$

$\\ $Proof of (2) is that for any two distinct integers$ \ m,n \ $then if$ \ mp = r_m + k_mq \ $and$ \ np = r_n + k_nq \ $then$ \ r_n \neq r_m \ $as it is the remainders that get mapped onto set of$ \ j$

$\\ $Suppose to the contrary that$ \ r_m = r_n \ $then$ \ \frac{(m-n)p}{q} = K \ $for some integer$ \ K$

$\\ $Given the definition of$ \ p,q \ $then$ \ p \ $and$ \ q \ $have no common factors, moreover, since$ \ 0 < m-n < q-1 \ $then$ \ (m-n) \ $is not divisible by$ \ q \ $then in order for$ \ \frac{(m-n)p}{q} = K \ $all the factors of$ \ q \ $must divide into either$ \ (m-n) \ $or$ \ p \ $but since it can't be$ \ p \ $then it must be$ \ (m-n)$

$\\ $But not all the factors of$ \ q \ $can divide into$ \ (m-n) \ $as that would mean$ \ (m-n) \ $is divisible by$ \ q \ $which is already shown to be impossible$$

$\\ $Therefore$ \ \frac{(m-n)p}{q} \neq K \ $thus$ \ r_m \neq r_n \ $thus$ \ (2) \ $is proven$$

$\\ $Thus the proposition in question is proven$$

-----------------

$\\ $NOTE: (*) means that at HSC level at least, this at most finds all$ \ \textbf{real} \ $polynomials with the aforementioned properties$$

Tell me if I'm going full weirdo mode but isn't this factorization only true if q is odd?

I'm talking about the part where you factorised (t+h)^q-t^q

Sy123 · Oct 29, 2015

Re: HSC 2016 4U Marathon - Advanced Level

lita1000 said:
Tell me if I'm going full weirdo mode but isn't this factorization only true if q is odd?

I'm talking about the part where you factorised (t+h)^q-t^q

You're thinking of the factorisation of (t+h)^q + t^q

$\\ 1 + x + \dots + x^{q-1} = \frac{x^q -1}{x-1}$

$\\ $Let$ \ x = \frac{t+h}{t} \ $and multiply both sides by$ \ t^{q-1}$

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