F 2

Art of Problem Solving

AoPS Online

PEN Math problems from the collection "Problems in Elementary Number Theory"

Math problems from the collection "Problems in Elementary Number Theory"

3 M G

BBookmark VNew Topic kLocked

PEN Math problems from the collection "Problems in Elementary Number Theory"

Math problems from the collection "Problems in Elementary Number Theory"

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

F 2

G H J

Reveal topic tags

rational numbers

L

G H BBookmark kLocked kLocked NReply

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

3615 posts

Peter

#1 h May 25, 2007, 12:24 AM • 2 Y

Y by Adventure10, Mango247

Find all $x$ and $y$ which are rational multiples of $\pi$ with $0<x<y<\frac{\pi}{2}$ and $\tan x+\tan y =2$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

667 posts

#2 h May 25, 2007, 12:24 AM • 2 Y

Y by Adventure10, Mango247

Let me complete the solution:
Firstly,if $\tan x+\tan y=2$ ,then
$\sin(x+y)=2\cos x\cos y=cos(x+y)+\cos(x-y)\Longrightarrow \sqrt{2}(\sin(x+y-\frac{\pi}{4}))=\cos(x-y) \Longrightarrow 2(\sin(x+y-\frac{\pi}{4}))^{2}=1-\sin^{2}(x-y) \Longrightarrow \cos(2(x+y)-\frac{\pi}{2})=\cos^{2}(x-y)$
Let $z=x-y,t=2(x+y)-\frac{\pi}{2}$
then $\cos^{2}z=\cos t$
Then we only need to find all pairs of (z,t) such that $z,t$ both are
rational multiples of $\pi$ and $\cos^{2}z=\cos t$

Lemma1,Define ${p_{n}(x)=\prod_{1\le k \le \frac{n}{2},gcd(k,n)=1}(x-\cos \frac{2k\pi}{n})},(n\ge 3)$ then we
firstly proved $p_{n}(x)$ is an polynomial with rational
coeffients,and it is irreducible(Do I spell correctly?I have not a
dictionary here...).

Proof of Lemma 1:Let $f_{n}(x)$ be the polynomial with the property $f_{n}(\cos x)=\cos nx$ .
then we can prove that $f_{2n+1}(x)-1=(x-1)h_{2n+1}(x)^{2},f_{2n}=(x-1)(x+1)h_{2n}(x)^{2}$
where $h_{n}(x)=\prod_{1\le k< \frac{n}{2}}{(x-\cos{\frac{2k\pi}{n})}}$
and we can easily find that $\prod_{d|n}{p_{d}(x)}=h_{n}(x)$
then $p_{n}(x)=\prod_{d|n}{(h_{d}(x))^{\mu(\frac{n}{d})}}$ is a polynomial.
We consider ${p_{n}(\frac{t+\frac{1}{t}}{2})=\prod_{1\le k \le \frac{n}{2},gcd(k,n)=1}(\frac{t+\frac{1}{t}}{2}-\cos \frac{2k\pi}{n})}=$ , $\frac{\Phi_{n}(t)}{t^{\frac{\phi(n)}{2}}}$
becuase $\Phi_{n}(t)$ is irreducible, $p_{n}(x)$ is also irreducible.

Lemma 2IF a polynomial $f(x)\in Q[x]$ is
irreducible, and $f(x^{2})$ is reducible,then there exist $g(x)\in Q[x]$ such that $f(x^{2})=\pm g(x)g(-x)$ .

Suppose that $z=\frac{u}{v}\pi ,t=\frac{m}{n}\pi,(n,v\ge 3)$
and Let $q(x)=p_{n}(x^{2})$ ,then $p_{v}(\cos z)=0,p_{n}(\cos t)=0,q(\cos z)=0$ by Lemma 1 , $p_{v}(x)|q(x)$ by Lemma 2, $p_{v}(x)=q(x) ,or, p_{v}(x)p_{v}(-x)=\pm q(x)$
But both of these two case leads to contraduction,becuase $q(x)$ has at least one non-real root,but $p_{v}(x)$ does not.

I hope my solution is correct.Could somebody tell me if it is correct?I am not so confident.......

@Peter,could you help me delet the first two post?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1144 posts

Tiks

#3 h May 25, 2007, 12:24 AM • 2 Y

Y by Adventure10, Mango247

Hawk Tiger wrote:

@Peter,could you help me delet the first two post?

I have deleted the posts :smile: ,though I am not Peter

.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

602 posts

hxy09

#4 h Feb 7, 2009, 9:49 AM • 2 Y

Y by Adventure10, Mango247

Hawk Tiger wrote:

Let me complete the solution:
Firstly,if $\tan x + \tan y = 2$ ,then
$\sin(x + y) = 2\cos x\cos y = cos(x + y) + \cos(x - y)\Longrightarrow \sqrt {2}(\sin(x + y - \frac {\pi}{4})) = \cos(x - y) \Longrightarrow 2(\sin(x + y - \frac {\pi}{4}))^{2} = 1 - \sin^{2}(x - y) \Longrightarrow \cos(2(x + y) - \frac {\pi}{2}) = \cos^{2}(x - y)$
Let $z = x - y,t = 2(x + y) - \frac {\pi}{2}$
then $\cos^{2}z = \cos t$
Then we only need to find all pairs of (z,t) such that $z,t$ both are
rational multiples of $\pi$ and $\cos^{2}z = \cos t$

Lemma1,Define ${ p_{n}(x) = \prod_{1\le k \le \frac {n}{2},gcd(k,n) = 1}(x - \cos \frac {2k\pi}{n})},(n\ge 3)$ then we
firstly proved $p_{n}(x)$ is an polynomial with rational
coeffients,and it is irreducible(Do I spell correctly?I have not a
dictionary here...).

Proof of Lemma 1:Let $f_{n}(x)$ be the polynomial with the property $f_{n}(\cos x) = \cos nx$ .
then we can prove that $f_{2n + 1}(x) - 1 = (x - 1)h_{2n + 1}(x)^{2},f_{2n} = (x - 1)(x + 1)h_{2n}(x)^{2}$
where $h_{n}(x) = \prod_{1\le k < \frac {n}{2}}{(x - \cos{\frac {2k\pi}{n})}}$
and we can easily find that $\prod_{d|n}{p_{d}(x)} = h_{n}(x)$
then $p_{n}(x) = \prod_{d|n}{(h_{d}(x))^{\mu(\frac {n}{d})}}$ is a polynomial.
We consider ${ p_{n}(\frac {t + \frac {1}{t}}{2}) = \prod_{1\le k \le \frac {n}{2},gcd(k,n) = 1}(\frac {t + \frac {1}{t}}{2} - \cos \frac {2k\pi}{n})} =$ , $\frac {\Phi_{n}(t)}{t^{\frac {\phi(n)}{2}}}$
becuase $\Phi_{n}(t)$ is irreducible, $p_{n}(x)$ is also irreducible.

Lemma 2IF a polynomial $f(x)\in Q[x]$ is
irreducible, and $f(x^{2})$ is reducible,then there exist $g(x)\in Q[x]$ such that $f(x^{2}) = \pm g(x)g( - x)$ .

Suppose that $z = \frac {u}{v}\pi ,t = \frac {m}{n}\pi,(n,v\ge 3)$
and Let $q(x) = p_{n}(x^{2})$ ,then $p_{v}(\cos z) = 0,p_{n}(\cos t) = 0,q(\cos z) = 0$ by Lemma 1 , $p_{v}(x)|q(x)$ by Lemma 2, $p_{v}(x) = q(x) ,or, p_{v}(x)p_{v}( - x) = \pm q(x)$
But both of these two case leads to contraduction,becuase $q(x)$ has at least one non-real root,but $p_{v}(x)$ does not.

I hope my solution is correct.Could somebody tell me if it is correct?I am not so confident.......

@Peter,could you help me delet the first two post?

Thank you for brilliant lemma 1.
BUT I don't think $Q(X) = \pm P_v(X)P_v( - X)$
Lemma ONLY tell you $P_v(X)P_v( - X)$ divides $Q(X)$
Maybe $Q(X) = P_v(X)P_v( - X)R(X)R( - X)$
Right?

This post has been edited 1 time. Last edited by hxy09, Feb 7, 2009, 1:07 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

3615 posts

Peter

#5 h Feb 7, 2009, 10:34 AM • 2 Y

Y by Adventure10, Mango247

hxy09 wrote:

Lemma ONLY tell you $P_v(X)P_v( - X)$ divides $Q(X)$
Maybe $Q(X) = P_v(X)P_v( - X)R(X)R( - X)$

Thenyou write $Q(X) = \big(P_v(X)R(X)\big)\big(P_v( - X)R( - X)\big)$ , so his claim for $g(X)=P_v(X)R(X)$ ? (Your claim misses the $\pm$ though)

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

602 posts

hxy09

#6 h Feb 7, 2009, 1:16 PM • 2 Y

Y by Adventure10, Mango247

Peter wrote:

hxy09 wrote:

Lemma ONLY tell you $P_v(X)P_v( - X)$ divides $Q(X)$
Maybe $Q(X) = P_v(X)P_v( - X)R(X)R( - X)$

Thenyou write $Q(X) = \big(P_v(X)R(X)\big)\big(P_v( - X)R( - X)\big)$ , so his claim for $g(X) = P_v(X)R(X)$ ? (Your claim misses the $\pm$ though)

Sorry I think you didn't really catch my question.Lemma 2 is obviously true.What I mean is $q(x)$ has at least one non real root but $p_v(x)$ doesn't.HOWEVER it doesn't lead to contradiction.You see, $R(x)$ and $R( - x)$ might have non real root.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

3615 posts

Peter

#7 h Feb 8, 2009, 10:58 AM • 2 Y

Y by Adventure10, Mango247

I'm not sure I understand you.

His argument (at least how I interpret it) is that $p_v$ is irreducible, $p_v|q$ , and $\deg(q)=2\deg p_v$ by definition. Hence if $g(x)=p_v(x)R(x)$ , then $\deg(R)=0$ ...

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

602 posts

hxy09

#8 h Feb 14, 2009, 8:38 AM • 2 Y

Y by Adventure10, Mango247

Peter wrote:

I'm not sure I understand you.

His argument (at least how I interpret it) is that $p_v$ is irreducible, $p_v|q$ , and $\deg(q) = 2\deg p_v$ by definition. Hence if $g(x) = p_v(x)R(x)$ , then $\deg(R) = 0$ ...

Sorry,but I think $\deg(q) = 2\deg p_n$ by definition not $\deg(q) = 2\deg p_v$ right?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

229 posts

#9 h Jul 13, 2016, 5:07 AM • 3 Y

Y by Williamgolly, Adventure10, Mango247

Solution:Take derivtive to find $tanx+tany$ is monotonic increasing as $x$ or $y$ increases.
So the only solution is $x=y=\frac {\pi}{4}$

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a