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jlacosta   0
May 1, 2025
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0 replies
jlacosta
May 1, 2025
0 replies
Simple inequality
sqing   6
N a minute ago by ys33
Source: Shiing-Shen Chern Cup Mathematical Olympiads 2018,Q2
Let $a,b,c,d$ be positive real numbers.Prove that$$\sqrt[3]{ab}+\sqrt[3]{cd}\leq\sqrt[3]{(a+b+c)(c+d+a)}.$$When equality holds?
6 replies
sqing
Jul 26, 2018
ys33
a minute ago
positive integers forming a perfect square
cielblue   4
N 2 minutes ago by Assassino9931
Find all positive integers $n$ such that $2^n-n^2+1$ is a perfect square.
4 replies
cielblue
May 2, 2025
Assassino9931
2 minutes ago
Interesting inequalities
sqing   2
N 3 minutes ago by pooh123
Source: Own
Let $ a,b> 0 $ and $  a^2+ab+b^2=a+b   $. Prove that
$$   \frac{a }{2b^2+1}+ \frac{b }{2a^2+1}+ \frac{1}{2ab+1} \geq \frac{21}{17}$$Let $ a,b> 0 $ and $ a^2+ab+b^2=a+b+1   $. Prove that
$$   \frac{a }{2b^2+1}+ \frac{b }{2a^2+1}+ \frac{1}{2ab+1} \geq1$$
2 replies
+1 w
sqing
May 4, 2025
pooh123
3 minutes ago
Labelling edges of Kn
oVlad   0
3 minutes ago
Source: Romania Junior TST 2025 Day 2 P3
Let $n\geqslant 3$ be an integer. Ion draws a regular $n$-gon and all its diagonals. On every diagonal and edge, Ion writes a positive integer, such that for any triangle formed with the vertices of the $n$-gon, one of the numbers on its edges is the sum of the two other numbers on its edges. Determine the smallest possible number of distinct values that Ion can write.
0 replies
1 viewing
oVlad
3 minutes ago
0 replies
No more topics!
number theory
Levieee   7
N Apr 19, 2025 by g0USinsane777
Idk where it went wrong, marks was deducted for this solution
$\textbf{Question}$
Show that for a fixed pair of distinct positive integers \( a \) and \( b \), there cannot exist infinitely many \( n \in \mathbb{Z} \) such that
\[
\sqrt{n + a} + \sqrt{n + b} \in \mathbb{Z}.
\]
$\textbf{Solution}$

Let
\[
x = \sqrt{n + a} + \sqrt{n + b} \in \mathbb{N}.
\]
Then,
\[
x^2 = (\sqrt{n + a} + \sqrt{n + b})^2 = (n + a) + (n + b) + 2\sqrt{(n + a)(n + b)}.
\]So:
\[
x^2 = 2n + a + b + 2\sqrt{(n + a)(n + b)}.
\]
Therefore,
\[
\sqrt{(n + a)(n + b)} \in \mathbb{N}.
\]
Let
\[
(n + a)(n + b) = k^2.
\]Assume \( n + a \neq n + b \). Then we have:
\[
n + a \mid k \quad \text{and} \quad k \mid n + b,
\]or it could also be that \( k \mid n + a \quad \text{and} \quad n + b \mid k \).

Without loss of generality, we take the first case:
\[
(n + a)k_1 = k \quad \text{and} \quad kk_2 = n + b.
\]
Thus,
\[
k_1 k_2 = \frac{n + b}{n + a}.
\]
Since \( k_1 k_2 \in \mathbb{N} \), we have:
\[
k_1 k_2 = 1 + \frac{b - a}{n + a}.
\]
For infinitely many \( n \), \( \frac{b - a}{n + a} \) must be an integer, which is not possible.

Therefore, there cannot be infinitely many such \( n \).
7 replies
Levieee
Apr 18, 2025
g0USinsane777
Apr 19, 2025
number theory
G H J
G H BBookmark kLocked kLocked NReply
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Levieee
220 posts
#1
Y by
Idk where it went wrong, marks was deducted for this solution
$\textbf{Question}$
Show that for a fixed pair of distinct positive integers \( a \) and \( b \), there cannot exist infinitely many \( n \in \mathbb{Z} \) such that
\[
\sqrt{n + a} + \sqrt{n + b} \in \mathbb{Z}.
\]
$\textbf{Solution}$

Let
\[
x = \sqrt{n + a} + \sqrt{n + b} \in \mathbb{N}.
\]
Then,
\[
x^2 = (\sqrt{n + a} + \sqrt{n + b})^2 = (n + a) + (n + b) + 2\sqrt{(n + a)(n + b)}.
\]So:
\[
x^2 = 2n + a + b + 2\sqrt{(n + a)(n + b)}.
\]
Therefore,
\[
\sqrt{(n + a)(n + b)} \in \mathbb{N}.
\]
Let
\[
(n + a)(n + b) = k^2.
\]Assume \( n + a \neq n + b \). Then we have:
\[
n + a \mid k \quad \text{and} \quad k \mid n + b,
\]or it could also be that \( k \mid n + a \quad \text{and} \quad n + b \mid k \).

Without loss of generality, we take the first case:
\[
(n + a)k_1 = k \quad \text{and} \quad kk_2 = n + b.
\]
Thus,
\[
k_1 k_2 = \frac{n + b}{n + a}.
\]
Since \( k_1 k_2 \in \mathbb{N} \), we have:
\[
k_1 k_2 = 1 + \frac{b - a}{n + a}.
\]
For infinitely many \( n \), \( \frac{b - a}{n + a} \) must be an integer, which is not possible.

Therefore, there cannot be infinitely many such \( n \).
This post has been edited 2 times. Last edited by Levieee, Apr 18, 2025, 7:47 PM
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Lil_flip38
53 posts
#2
Y by
I might be missing something, but why does \((n+a)(n+b)=k^2\) imply \(n+a\mid k\) and \(k\mid n+b\)? Why cant for example \(n+a=25\) and \(n+b=4\) and \(k=10\). The factors of \(k\) do not have to be distributed nicely between \(n+a\) and \(n+b\).
Z K Y
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DTforever
5 posts
#3
Y by
I think the condition implies both $n+a$ and $n+b$ are perfect squares.
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MathLuis
1523 posts
#4 • 1 Y
Y by Levieee
You need a slightly more sophisitcated finish after realising $(n+a)(n+b)$ is a square for infinitely many $n$ and it is that we have $n^2+(a+b)n+ab$ is a square for infinitely many $n$ and now setting some arbitrarily large $n$ and considering some $k$ for which $2k \ge a+b >2k-2$ we could have that it is a square bounded between $(n+k)^2$ and $(n+k-1)^2$ so it has to be $(n+k)^2$ which shows that $(a+b)^2=4ab$ by coefficient checking and matching and thus $(a-b)^2=0$ so $a=b$ which is a contradiction!.
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Safal
169 posts
#5
Y by
Levieee wrote:
Idk where it went wrong, marks was deducted for this solution
$\textbf{Question}$
Show that for a fixed pair of distinct positive integers \( a \) and \( b \), there cannot exist infinitely many \( n \in \mathbb{Z} \) such that
\[
\sqrt{n + a} + \sqrt{n + b} \in \mathbb{Z}.
\]
$\textbf{Solution}$

Let
\[
x = \sqrt{n + a} + \sqrt{n + b} \in \mathbb{N}.
\]
Then,
\[
x^2 = (\sqrt{n + a} + \sqrt{n + b})^2 = (n + a) + (n + b) + 2\sqrt{(n + a)(n + b)}.
\]So:
\[
x^2 = 2n + a + b + 2\sqrt{(n + a)(n + b)}.
\]
Therefore,
\[
\sqrt{(n + a)(n + b)} \in \mathbb{N}.
\]
Let
\[
(n + a)(n + b) = k^2.
\]Assume \( n + a \neq n + b \). Then we have:
\[
n + a \mid k \quad \text{and} \quad k \mid n + b,
\]or it could also be that \( k \mid n + a \quad \text{and} \quad n + b \mid k \).

Without loss of generality, we take the first case:
\[
(n + a)k_1 = k \quad \text{and} \quad kk_2 = n + b.
\]
Thus,
\[
k_1 k_2 = \frac{n + b}{n + a}.
\]
Since \( k_1 k_2 \in \mathbb{N} \), we have:
\[
k_1 k_2 = 1 + \frac{b - a}{n + a}.
\]
For infinitely many \( n \), \( \frac{b - a}{n + a} \) must be an integer, which is not possible.

Therefore, there cannot be infinitely many such \( n \).

$k_1$ and $k_2$ need not be integers, they can be rational beacuse we can have that the denominator ot that rational divide $n+a$
This post has been edited 1 time. Last edited by Safal, Apr 18, 2025, 9:19 PM
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Levieee
220 posts
#6
Y by
Safal wrote:
Levieee wrote:
Idk where it went wrong, marks was deducted for this solution
$\textbf{Question}$
Show that for a fixed pair of distinct positive integers \( a \) and \( b \), there cannot exist infinitely many \( n \in \mathbb{Z} \) such that
\[
\sqrt{n + a} + \sqrt{n + b} \in \mathbb{Z}.
\]
$\textbf{Solution}$

Let
\[
x = \sqrt{n + a} + \sqrt{n + b} \in \mathbb{N}.
\]
Then,
\[
x^2 = (\sqrt{n + a} + \sqrt{n + b})^2 = (n + a) + (n + b) + 2\sqrt{(n + a)(n + b)}.
\]So:
\[
x^2 = 2n + a + b + 2\sqrt{(n + a)(n + b)}.
\]
Therefore,
\[
\sqrt{(n + a)(n + b)} \in \mathbb{N}.
\]
Let
\[
(n + a)(n + b) = k^2.
\]Assume \( n + a \neq n + b \). Then we have:
\[
n + a \mid k \quad \text{and} \quad k \mid n + b,
\]or it could also be that \( k \mid n + a \quad \text{and} \quad n + b \mid k \).

Without loss of generality, we take the first case:
\[
(n + a)k_1 = k \quad \text{and} \quad kk_2 = n + b.
\]
Thus,
\[
k_1 k_2 = \frac{n + b}{n + a}.
\]
Since \( k_1 k_2 \in \mathbb{N} \), we have:
\[
k_1 k_2 = 1 + \frac{b - a}{n + a}.
\]
For infinitely many \( n \), \( \frac{b - a}{n + a} \) must be an integer, which is not possible.

Therefore, there cannot be infinitely many such \( n \).

$k_1$ and $k_2$ need not be integers, they can be rational beacuse we can have that the denominator ot that rational divide $n+a$

yea but after it's reduced we can get it to be an integer , doesnt $ a \mid b$ imply for some $k$ $ak=b$ , k is an integer

@2above how do u say that $n^2 + ab + (a+b)n$
This post has been edited 2 times. Last edited by Levieee, Apr 19, 2025, 7:41 AM
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Levieee
220 posts
#7
Y by
DTforever wrote:
I think the condition implies both $n+a$ and $n+b$ are perfect squares.

does the counter example provided by @Lil_flip38 work? or that counter example can be invalidated because $n = 4-b$, $b \ge 1$ so as n increases we can just show it cant happen
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g0USinsane777
48 posts
#8 • 1 Y
Y by Levieee
Levieee wrote:
Assume \( n + a \neq n + b \). Then we have:
\[
n + a \mid k \quad \text{and} \quad k \mid n + b,
\]or it could also be that \( k \mid n + a \quad \text{and} \quad n + b \mid k \).
This part is certainly wrong, since $(n+a)(n+b)=k^2$ can imply $n+a \leq k$ and $n+b \geq k$ or vice-versa but not the divisibility condition above. You can find many counter examples to this.

Another finish from this part can be that let $\gcd(n+a,n+b)=d$. (of course, $n+a \neq n+b$)
So, $n+a = dx$ and $n+b=dy$, $\gcd(x,y)=1$
This gives us that, $d^2xy = k^2$
Which means that $xy=l^2$ and $x=u^2$ and $y=v^2$, $\gcd(u,v)=1$.
This means that, $b-a=d(v^2 - u^2)$ for a particular $n$.
Since, the original condition needs to be true for infinitely many $n$, by infinite PHP we will have infinitely many $n$ for which we get the same value of $d$ for a particular value of $k$.
So, for infinitely many $n$, which in turn give infinitely many value of the pair $(u,v)$, the value $\frac{b-a}{d}$ is constant, say $t$.
From the above argument, since infinitely many pairs $(u,v)$ satisfy $t = v^2-u^2 = (v-u)(v+u)$, where $v,u$ are integers, this means that $t$ has infinitely many factors $\implies$ $t=0$, i.e., $b=a$ which contradicts that $a$ and $b$ are distinct.
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