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Solve a system of integers
nhathhuyyp5c   0
19 minutes ago
Find all triplets $a,b,c,d$ where $a,b,c,d$ are all integers and $abcd\neq 0$ such that $$\begin{cases} a^2+b^2+c^2+d^2\leq 400.\\ a^3+b^3+c^3+d^3\geq 4000. \\ a^4+b^4+c^4+d^4\leq 40001. \end{cases}$$
0 replies
nhathhuyyp5c
19 minutes ago
0 replies
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Calculating combinatorial numbers
lgx57   5
N 3 hours ago by generatingFraction
Try to simplify this expression:

$$\sum_{i=1}^n \sum_{j=1}^i C_{n}^i C_{n}^j$$
5 replies
lgx57
4 hours ago
generatingFraction
3 hours ago
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Proper subsets of R
lgx57   0
4 hours ago
Let $S_1,S_2 \cdots S_n$ are proper subsets of $\mathbb{R}$ and they are closed for addition and subtraction. Try to prove that:

$$\displaystyle\bigcup_{i=1}^n S_i \ne \mathbb{R}$$
0 replies
lgx57
4 hours ago
0 replies
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Rubik's cube problem
ilikejam   18
N Today at 5:15 AM by mpcnotnpc
If I have a solved Rubik's cube, and I make a finite sequence of (legal) moves repeatedly, prove that I will eventually resolve the puzzle.

(this wording is kinda goofy but i hope its sorta intuitive)
18 replies
ilikejam
Mar 28, 2025
mpcnotnpc
Today at 5:15 AM
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line AO passes through the midpoint of segment EF
toanrathay   0
Today at 5:08 AM
Given a triangle \( ABC \) with \( AB < AC \) and the angle bisector \( AD \).
The line passing through \( A \) and perpendicular to \( AC \) intersects the line passing through \( B \) and parallel to \( AD \) at point \( E \).
The line passing through \( A \) and perpendicular to \( AB \) intersects the line passing through \( C \) and parallel to \( AD \) at point \( F \).
Let \( O \) be the intersection of the three perpendicular bisectors of triangle \( ABC \).
Prove that line \( AO \) passes through the midpoint of segment \( EF \).
0 replies
toanrathay
Today at 5:08 AM
0 replies
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Inequalities
sqing   1
N Today at 4:16 AM by sqing
Let $ a,b\geq 0 $ and $a^2+b^2+ab+a+b=1. $ Prove that
$$\frac{179657}{450000}\geq a^2+b^2+3ab(a+ b-0.13562)\geq \frac{3-\sqrt 5}{2}$$
1 reply
sqing
Today at 4:03 AM
sqing
Today at 4:16 AM
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Weird fractions
wangyanliluke   1
N Today at 3:27 AM by Facejo
While I was doing a question I made this really weird observation:

So first, we suppose $S$ is the infinite sum $\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...$. Then $S$ is more than $0$ since $\frac{1}{1}>\frac{1}{2}$, $\frac{1}{3}>\frac{1}{4}$, and so on. But we can rewrite it as $\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...-2(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...)=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...-(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...)=0.$ So is $S$ more than $0$ or equal to $0$? Help is much appreciated
1 reply
wangyanliluke
Today at 3:15 AM
Facejo
Today at 3:27 AM
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Prove that \( S \) contains all integers.
nhathhuyyp5c   1
N Today at 3:19 AM by GreenTea2593
Let \( S \) be a set of integers satisfying the following property: For every positive integer \( n \) and every set of coefficients \( a_0, a_1, \dots, a_n \in S \), all integer roots of the polynomial $P(x) = a_0 + a_1 x + \dots + a_n x^n $ are also elements of \( S \). It is given that \( S \) contains all numbers of the form \( 2^a - 2^b \) where \( a, b \) are positive integers. Prove that \( S \) contains all integers.









1 reply
nhathhuyyp5c
Yesterday at 3:53 PM
GreenTea2593
Today at 3:19 AM
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Cool one
MTA_2024   11
N Today at 2:42 AM by sqing
Prove that for all real numbers $a$ and $b$ verifying $a>b>0$ . $$(n+1) \cdot b^n \leq \frac{a^{n+1}-b^{n+1}}{a-b} \leq (n+1) \cdot a^n $$
11 replies
MTA_2024
Mar 15, 2025
sqing
Today at 2:42 AM
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4-digit evens with 0,1,2,3,4,5 (Puerto Rico TST 2024.6)
Equinox8   3
N Yesterday at 11:59 PM by wisewigglyjaguar
Find the sum of all $4$-digit even numbers that can be written using the digits $0, 1, 2, 3, 4,$ and $5$. Digits can be repeated in a number.
3 replies
Equinox8
Mar 24, 2025
wisewigglyjaguar
Yesterday at 11:59 PM
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Help me please
Hahahsafdk3l   2
N Yesterday at 11:43 PM by Hahahsafdk3l
Find all functions $f: \mathbb{R^+} \to \mathbb{R^+}$ such that $f(2f(x) + f(y) + xyy) = xy + 2x + y, \forall x, y > 0$
2 replies
Hahahsafdk3l
Yesterday at 2:21 AM
Hahahsafdk3l
Yesterday at 11:43 PM
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Mathematics
Moumi   7
N Apr 18, 2016 by Virgil Nicula
Definition domain of
Rad(x+rad(x^4+3))
Rad= radical
7 replies
Moumi
Feb 9, 2016
Virgil Nicula
Apr 18, 2016
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Moumi
5 posts
Moumi
#1 h  Feb 9, 2016, 12:48 PM • 1 Y
Y by Adventure10
Definition domain of
Rad(x+rad(x^4+3))
Rad= radical
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ythomashu
6322 posts
ythomashu
#2 h  Feb 9, 2016, 1:31 PM • 3 Y
Y by Moumi, Adventure10, Mango247
$\sqrt{x+\sqrt{x^4+3}}$
All reals
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Moumi
5 posts
Moumi
#3 h  Feb 9, 2016, 1:42 PM • 2 Y
Y by Adventure10, Mango247
Can i have the explanation please.thank you
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mathguy5041
2659 posts
mathguy5041
#4 h  Feb 9, 2016, 1:44 PM • 2 Y
Y by Adventure10, Mango247
Why can't $x+\sqrt{x^4+3}$ be negative?
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ythomashu
6322 posts
ythomashu
#5 h  Feb 9, 2016, 1:49 PM • 2 Y
Y by Adventure10, Mango247
$x+\sqrt{x^4+3}=0$ I think it has no real solution.
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HVishy
312 posts
HVishy
#6 h  Feb 9, 2016, 2:34 PM • 2 Y
Y by Adventure10, Mango247
ythomashu wrote:
$\sqrt{x+\sqrt{x^4+3}}$
All reals

As x approaches positive or negative infinity, the $\sqrt{x^4+3}$ will become $x^2$.

We know that for larger negative or positive cases, $x^2+x$ is always positive.

As for the smaller cases, a quick check shows the function increasing over the positive numbers.
This post has been edited 2 times. Last edited by HVishy, Feb 9, 2016, 3:45 PM
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brokenfixer
53 posts
brokenfixer
#7 h  Mar 9, 2016, 1:19 AM • 1 Y
Y by Adventure10
mathguy5041 wrote:
Why can't $x+\sqrt{x^4+3}$ be negative?

$(a-1)^2 \ge 0$

$a^2 -2a +1 \ge 0$

$2a \le a^2+1$

$a \le \frac{1}{2}(a^2+1) < (\frac{1}{2}a^2 + \frac{1}{2}) + (\frac{5}{2}) + (\frac{1}{2}a^2)$

So it is pretty clear that we always have
$a < a^2+3$

Using $a = x^2$
$x^2 < x^4 + 3$

$\sqrt{x^2} < \sqrt{x^4+3}$

$|x| < \sqrt{x^4+3}$

Which means that your expression can't become negative.
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Virgil Nicula
7054 posts
Virgil Nicula
#8 h  Apr 18, 2016, 3:24 PM • 2 Y
Y by Adventure10, Mango247
Denote $f(x)=x+\sqrt{x^4+3}\ ,\ x\in\mathbb D_f$. Observe that $(\forall )\ x\ge 0$ the function $f$ is strict increasing $(\nearrow )$ and $f(x)\ge \sqrt 3$ . Now can suppose that $x<0$ . Hence $1-x> 1>0$ and $\sqrt{x^4+3}\ge 1-x\iff$ $x^4+3\ge (1-x)^2\iff$ $x^4-x^2+2x+2\ge 0\iff$ $(x+1)^2\left(x^2-2x+2\right)\ge 0$ what is truly. Therefore, $(\forall )\ x<0\ ,\ x+\sqrt{x^4+3}\ge 1$ and $f(-1)=1$ . In conclusion, $\boxed{(\forall )\ x\in\mathbb R\ ,\ f(x)\ge 1=f(-1)\ ,\ \mathrm{i.e.}\ f:\mathbb R\rightarrow [1,\infty)}$ and $\mathrm {Im(f)}=f(\mathbb R)=[1,\infty ).$ See here and the proposed problem PP6 from here.
This post has been edited 16 times. Last edited by Virgil Nicula, Apr 19, 2016, 9:39 PM
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