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0 replies
jlacosta
Mar 2, 2025
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Peter   4
N Thursday at 3:29 AM by sansgankrsngupta
Let $ \alpha$ be a rational number with $ 0 < \alpha < 1$ and $ \cos (3 \pi \alpha) + 2\cos(2 \pi \alpha) = 0$. Prove that $ \alpha = \frac {2}{3}$.
4 replies
Peter
May 25, 2007
sansgankrsngupta
Thursday at 3:29 AM
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Peter   13
N Mar 2, 2025 by EVS383
Prove no three lattice points in the plane form an equilateral triangle.
13 replies
Peter
May 25, 2007
EVS383
Mar 2, 2025
J
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A 22
Peter   6
N Sep 21, 2024 by pie854
Prove that the number \[\sum_{k=0}^{n}\binom{2n+1}{2k+1}2^{3k}\] is not divisible by $5$ for any integer $n\geq 0$.
6 replies
Peter
May 25, 2007
pie854
Sep 21, 2024
J
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A 30
Peter   10
N Sep 21, 2024 by pie854
Show that if $n \ge 6$ is composite, then $n$ divides $(n-1)!$.
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Peter
May 25, 2007
pie854
Sep 21, 2024
J
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A 21
Peter   11
N Sep 21, 2024 by pie854
Let n be a positive integer. Show that the product of $ n$ consecutive positive integers is divisible by $ n!$
11 replies
Peter
May 25, 2007
pie854
Sep 21, 2024
J
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Peter   11
N Sep 21, 2024 by pie854
Determine all positive integers $n$ for which there exists an integer $m$ such that $2^{n}-1$ divides $m^{2}+9$.
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Peter
May 25, 2007
pie854
Sep 21, 2024
J
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A 17
Peter   13
N Sep 19, 2024 by pie854
Let $m$ and $n$ be natural numbers such that \[A=\frac{(m+3)^{n}+1}{3m}\] is an integer. Prove that $A$ is odd.
13 replies
Peter
May 25, 2007
pie854
Sep 19, 2024
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Peter   10
N Sep 19, 2024 by pie854
Let $m$ and $n$ be natural numbers and let $mn+1$ be divisible by $24$. Show that $m+n$ is divisible by $24$.
10 replies
Peter
May 25, 2007
pie854
Sep 19, 2024
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Peter   10
N Sep 17, 2024 by pie854
Suppose that $k \ge 2$ and $n_{1}, n_{2}, \cdots, n_{k}\ge 1$ be natural numbers having the property \[n_{2}\; \vert \; 2^{n_{1}}-1, n_{3}\; \vert \; 2^{n_{2}}-1, \cdots, n_{k}\; \vert \; 2^{n_{k-1}}-1, n_{1}\; \vert \; 2^{n_{k}}-1.\] Show that $n_{1}=n_{2}=\cdots=n_{k}=1$.
10 replies
Peter
May 25, 2007
pie854
Sep 17, 2024
J
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A 14
Peter   9
N Sep 17, 2024 by pie854
Let $n$ be an integer with $n \ge 2$. Show that $n$ does not divide $2^{n}-1$.
9 replies
Peter
May 25, 2007
pie854
Sep 17, 2024
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Peter
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Peter
#1 h  May 25, 2007, 12:25 AM • 4 Y
Y by W.Sun, Adventure10, Mango247, and 1 other user
Prove no three lattice points in the plane form an equilateral triangle.
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mathmanman
1444 posts
mathmanman
#2 h  May 25, 2007, 12:25 AM • 3 Y
Y by Adventure10, Mango247, and 1 other user
The coordinates of the three vertices are :
$(x_0, y_0), (x_1 = x_0 + c \cos(t), y_1 = y_0 + c \sin(t))$ and $(x_2 = x_0 + c \cos(t+ \frac {\pi}3), y_2 = y_0 + c \sin(t+\frac {\pi}3)).$

Then :
If $y_1 \neq y_0 : \sqrt{3} = \frac {x_0 + x_1 - 2x_2}{y_1 - y_0} \implies$ impossible;
If $y_1 = y_0 : x_1 \neq x_0$ and $\sqrt{3} = 2 \frac {y_2 - y_0}{x_1 - x_0} \implies$ impossible as well.

So we're done.
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ideahitme
91 posts
ideahitme
#3 h  Aug 29, 2007, 7:43 AM • 3 Y
Y by Adventure10, Mango247, and 1 other user
Peter wrote:
Prove no three lattice points in the plane form an equilateral triangle.

We generalize it slightly.


Theorem. No three rational points in the plane form an equilateral triangle.


First Proof. Assume on the contrary that three vertices of an equilateral triangle $ ABC$ are rational points. Write $ A(a_{1}, b_{1})$, $ B(a_{2}, b_{2})$, $ C(a_{3}, b_{3})$, where $ a_{1}, b_{1}, a_{2}, b_{2}, a_{3}, b_{3}\in\mathbb{Q}$. Since $ ABC$ is equilateral, we obtain
\[ \left( a_{1}-a_{2}\right)^{2}+\left( b_{1}-b_{2}\right)^{2}=\left( a_{2}-a_{3}\right)^{2}+\left( b_{2}-b_{3}\right)^{2}=\left( a_{1}-a_{3}\right)^{2}+\left( b_{1}-b_{3}\right)^{2}=\lambda.\]
for some $ \lambda\in\mathbb{Q}$. After setting $ p=a_{1}-a_{2}$, $ q=a_{2}-a_{3}$, $ r=b_{1}-b_{2}$ and $ s=b_{2}-b_{3}$, it becomes
\[ p^{2}+r^{2}= q^{2}+s^{2}=\left( p+q\right)^{2}+\left( r+s\right)^{2}=\lambda.\]
We obtain
\[ 2 ( pq+rs ) =\left( p+q\right)^{2}+\left( r+s\right)^{2}-\left( p^{2}+r^{2}\right)-\left( q^{2}+s^{2}\right) =-\lambda.\]
or
\[ pq+rs =-\frac{1}{2}\lambda.\]
It follows that
\[ (ps-qr)^{2}=\left( p^{2}+r^{2}\right)\left( q^{2}+s^{2}\right)-( pq+rs)^{2}=\frac{3}{4}{\lambda}^{2}= 3\left(-\frac{\lambda}{2}\right)^{2}\]
or
\[ \vert ps-qr\vert =\sqrt{3}\,\vert pq+rs\vert.\]
Notice that both $ ps-qr$ and $ pq+rs$ are rational numbers. If $ pq+rs\neq 0$, then $ \sqrt{3}=\left\vert\frac{ps-qr}{pq+rs}\right\vert\in\mathbb{Q}$, which contradicts for the irrationality of $ \sqrt{3}$. Hence, $ pq+rs=0$. This implies that $ ps-qr=0$ so that
\[ \left( p^{2}+r^{2}\right)\left( q^{2}+s^{2}\right) = (pq+rs)^{2}+(ps-qr)^{2}=0.\]
Hence, we have $ (p,r)=(0,0)$ or $ (q,s)=(0,0)$. This means that
\[ A\left( a_{1}, b_{1}\right) = B\left( a_{2}, b_{2}\right)\;\; or\;\; B\left( a_{2}, b_{2}\right) = C\left( a_{3}, b_{3}\right).\]
Since $ ABC$ forms a triangle, this is a contradiction.

Second Proof. We need the following well-known result.

Lemma. Any triangle whose vertices have rational coordinates has rational area.

Proof of Lemma. Suppose that rational points $ P(a_{1}, b_{1})$, $ Q(a_{2}, b_{2})$, $ R(a_{3}, b_{3})$ form a triangle. By the well-known formula, we obtain
\[ [\triangle PQR ]=\frac{1}{2}\vert a_{1}b_{2}+a_{2}b_{3}+a_{3}b_{1}-a_{2}b_{1}-a_{3}b_{2}-a_{1}b_{2}\vert\in\mathbb{Q}.\]
Assume on the contrary that three rational points form an equilateral triangle with the length $ l$. Since vertices are rational points, we find that $ l^{2}$ is also positive rational. On the one hand, the above proposition means that the area of the triangle is rational. On the other hand, the area of triangle is equal to $ \frac{\sqrt{3}}{4}l^{2}$, which is irrational. This is a contradiction.

Third Proof. We present another alternative proof of Theorem 10.2. Assume on the contrary that three rational points form an equilateral triangle $ ABC$. Then, we may translate $ ABC$ to $ OB'C'$ by $ -\overrightarrow{OA}$, where $ O$ is the origin. Then, three rational points $ O$, $ B'(p,q)$, $ C'(r,s)$ also form an equilateral triangle. By $ \frac{\pi}{3}$ rotation around the origin, $ B'$ corresponds to $ C'$. In case of $ \frac{\pi}{3}$ \textit{counterclockwise} rotation, we obtain
\[ \left(\begin{array}{c}r\\ s\end{array}\right) =\left(\begin{array}{cc}\cos\frac{\pi}{3}&-\sin\frac{\pi}{3}\\ \sin\frac{\pi}{3}&\cos\frac{\pi}{3}\end{array}\right)\left(\begin{array}{c}p\\ q\end{array}\right)\]
or
\[ r =\frac{1}{2}p-\frac{\sqrt{3}}{2}q,\; s=\frac{\sqrt{3}}{2}p+\frac{1}{2}q.\]
Since $ r$ and $ p$ are rational numbers, we see that $ \frac{\sqrt{3}}{2}q=r-\frac{1}{2}p$ is rational. Since $ \frac{\sqrt{3}}{2}$ is irrational and since $ q$ is rational, this means that $ q=0$. From $ s=\frac{\sqrt{3}}{2}p+\frac{1}{2}q$, we also get $ p=0$. This means that $ B'$ coincides with the origin, which is a contradiction.
In case of $ \frac{\pi}{3}$ \textit{clockwise} rotation, you may change the role of $ B'$ and $ C'$ to reach the contradiction.
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pohoatza
1145 posts
pohoatza
#4 h  Apr 2, 2008, 11:19 AM • 3 Y
Y by Adventure10 and 2 other users
Here is a different approach leading to another generalization:
Assume that triangle $ ABC$ is a lattice equilateral triangle. It is known that if $ \omega$ is the Brocard angle of a triangle $ ABC$, then $ \cot{\omega} = \frac {a^{2} + b^{2} + c^{2}}{4S}$. By Pick's theorem, the area of triangle $ ABC$ is (at least) rational. Thus, $ \cot{\omega} \in \mathbb{Q}$. But since $ ABC$ is equilateral, its Brocard angle is $ \frac {\pi}{6}$, and therefore $ \cot{\omega} = \cot{\frac {\pi}{6}} = \sqrt {3} \not\in \mathbb{Q}$. Contradiction.

Generalization (communicated by Cezar Lupu): The Brocard angle of a lattice triangle is not a rational multiple of $ \pi$.
This now follows from the fact that $ \cot{\mathbb{Q}\pi} \in \left\{ - 1, 0, 1\right\}$, combined with Johnson's theorem (which says that the Brocard angle of a triangle is less or equal $ \pi/6$, with equality when $ ABC$ is equilateral - see here: http://mathworld.wolfram.com/BrocardAngle.html).
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pohoatza
1145 posts
pohoatza
#5 h  Apr 22, 2008, 12:56 PM • 2 Y
Y by Adventure10, Mango247
ideahitme wrote:
Theorem. No three rational points in the plane form an equilateral triangle.
See here for a possible generalization (in case that the second part is true).
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ZetaX
7579 posts
ZetaX
#6 h  Apr 22, 2008, 1:09 PM • 5 Y
Y by Adventure10, Mango247, Mango247, and 2 other users
There is also an n-dimensional version:

The vertices of an n-dimensional regular simplex can be realised as subset of $ \mathbb Q^n$ (respectively $ \mathbb Z^n$) iff $ n$ fulfills one of the following conditions:
- $ n \equiv 1 \mod 4$ and $ n + 1$ is a sum of two perfect squares.
- $ n \equiv 3 \mod 4$
- $ n$ is even and $ n + 1$ is a perfect square.
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pohoatza
1145 posts
pohoatza
#7 h  Apr 22, 2008, 2:05 PM • 2 Y
Y by Adventure10, Mango247
Nice! Though I have no idea how to prove it. Can you please tell me a reference or something? (It seems that I'm losing my Googling skills).
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polskimisiek
293 posts
polskimisiek
#8 h  Apr 22, 2008, 7:21 PM • 2 Y
Y by Adventure10, Mango247
Well, there is one more beautiful problem related to the topic.
Prove that for $ n\geq 3$ there is no n-gon with all sides equal where each of the vertex lies in the lattice points. (of course without n=4)
I'd like to see your ideas :wink:

Small hint
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ZetaX
7579 posts
ZetaX
#9 h  Apr 22, 2008, 7:25 PM • 3 Y
Y by Adventure10, Mango247, Mango247
In fact, the only $ n$-gons contained in any lattice are $ n=1,2,3,4,6$.

@pohoatza: it was proven by Schoenberg, you may be able to find something about this on the net.
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polskimisiek
293 posts
polskimisiek
#10 h  Apr 22, 2008, 7:28 PM • 2 Y
Y by Adventure10, Mango247
ZetaX wrote:
In fact, the only $ n$-gons contained in any lattice are $ n = 1,2,3,4,6$.
I don't quite understand what you meant by that :huh:
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ZetaX
7579 posts
ZetaX
#11 h  Apr 22, 2008, 7:46 PM • 1 Y
Y by Adventure10
The only $ n$ such that the regular $ n$-gon can be realized by points of a lattice (of any kind) are $ n=1,2,3,4,6$.
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pohoatza
1145 posts
pohoatza
#12 h  Apr 22, 2008, 8:44 PM • 2 Y
Y by Adventure10, Mango247
ZetaX wrote:
The only $ n$ such that the regular $ n$-gon can be realized by points of a lattice (of any kind) are $ n = 1,2,3,4,6$.
Actually (in the spirit of this topic) $ n \neq 3$. :D
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ZetaX
7579 posts
ZetaX
#13 h  Apr 22, 2008, 8:59 PM • 2 Y
Y by Adventure10, Mango247
A lattice is not necessarily 2-dimensional.
Another thing is that in my post, "lattice" can be a general lattice (a discrete subgroup of some $ \mathbb R^m$); e.g. we could have the hexagonal lattice.
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EVS383
18 posts
EVS383
#14 h  Mar 2, 2025, 5:19 PM
Y by
The area is rational by picks' theorem but irrational by A = s^2sqrt3/4
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