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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
inequality thing
BinariouslyRandom   2
N 7 minutes ago by BinariouslyRandom
Source: Philippine MO 2025 P5
Find the largest real constant $k$ for which the inequality \[ (a^2+3)(b^2+3)(c^2+3)(d^2+3) + k(a-1)(b-1)(c-1)(d-1) \ge 0 \]holds for all real numbers $a$, $b$, $c$, and $d$.

answer
2 replies
BinariouslyRandom
2 hours ago
BinariouslyRandom
7 minutes ago
Geometry Concurrence
KHOMNYO2   0
24 minutes ago
Given triangle $XYZ$ such that $XY \neq XZ$. Let excircle-$X$ be tangent with $YZ, ZX, XY$ at points $U, V, W$ respectively. Let $R$ and $S$ be points on the segment $XZ, XY$ respectively such that $RS$ is parallel to $YZ$. Lastly, let $\gamma$ be the circle that is externally tangent with the excircle-$X$ on point $T$. Prove that $VW, UT$, and $RS$ concur at a point.
0 replies
KHOMNYO2
24 minutes ago
0 replies
inequality
mathematical-forest   1
N 24 minutes ago by lbh_qys
Positive real numbers $x_{1} ,x_{2} \cdots ,x_{n}$,satisfied $\sum_{i=1}^{n}x_{i} =1$
Proof:$$\sum_{i=1}^{n} \frac{\min  \left \{  x_{i-1},x_{i}\right \}\max \left \{  x_{i},x_{i+1}\right \}  }{x_{i}} \le 1$$
1 reply
mathematical-forest
an hour ago
lbh_qys
24 minutes ago
Some number theory
EeEeRUT   5
N an hour ago by shafikbara48593762
Source: Thailand MO 2025 P9
Let $p$ be an odd prime and $S = \{1,2,3,\dots, p\}$
Assume that $U: S \rightarrow S$ is a bijection and $B$ is an integer such that $$B\cdot U(U(a)) - a \: \text{ is a multiple of} \: p \: \text{for all} \: a \in S$$Show that $B^{\frac{p-1}{2}} -1$ is a multiple of $p$.
5 replies
EeEeRUT
May 14, 2025
shafikbara48593762
an hour ago
No more topics!
Two Integer Sequences
Brut3Forc3   12
N Apr 13, 2023 by S.Das93
Source: USAMO 1973
Let $ \{X_n\}$ and $ \{Y_n\}$ denote two sequences of integers defined as follows:
\begin{align*} X_0 = 1,\ X_1 = 1,\ X_{n + 1} = X_n + 2X_{n - 1} \quad (n = 1,2,3,\ldots), \\
Y_0 = 1,\ Y_1 = 7,\ Y_{n + 1} = 2Y_n + 3Y_{n - 1} \quad (n = 1,2,3,\ldots).\end{align*}Prove that, except for the "1", there is no term which occurs in both sequences.
12 replies
Brut3Forc3
Mar 7, 2010
S.Das93
Apr 13, 2023
Two Integer Sequences
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G H BBookmark kLocked kLocked NReply
Source: USAMO 1973
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Brut3Forc3
1948 posts
#1 • 2 Y
Y by Adventure10, Mango247
Let $ \{X_n\}$ and $ \{Y_n\}$ denote two sequences of integers defined as follows:
\begin{align*} X_0 = 1,\ X_1 = 1,\ X_{n + 1} = X_n + 2X_{n - 1} \quad (n = 1,2,3,\ldots), \\
Y_0 = 1,\ Y_1 = 7,\ Y_{n + 1} = 2Y_n + 3Y_{n - 1} \quad (n = 1,2,3,\ldots).\end{align*}Prove that, except for the "1", there is no term which occurs in both sequences.
This post has been edited 2 times. Last edited by djmathman, Dec 20, 2016, 9:18 PM
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ocha
955 posts
#2 • 3 Y
Y by rayfish, Adventure10, Mango247
Assuming you actually mean $ X_{n+1}=X_n + 2X_{n-1}$ and the same for $ Y_n$ then,

taking the sequences $ \mod 8$

$ \{X_i\} = \{1,1,3, - 3,3, - 3,3,...\}$

$ \{Y_i\} = \{1, - 1,1, - 1,1, - 1,...\}$
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Brut3Forc3
1948 posts
#3 • 2 Y
Y by Adventure10, Mango247
ocha wrote:
Assuming you actually mean $ X_{n + 1} = X_n + 2X_{n - 1}$ and the same for $ Y_n$ then,

taking the sequences $ \mod 8$

$ \{X_i\} = \{1,1,3, - 3,3, - 3,3,...\}$

$ \{Y_i\} = \{1, - 1,1, - 1,1, - 1,...\}$
Yes, it was a typo. My bad.
How does this finish the problem?
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ocha
955 posts
#4 • 2 Y
Y by Adventure10, Mango247
Brut3Forc3 wrote:
How does this finish the problem?

if $ X_a=Y_b$ then we would have $ \pm 3\equiv \pm 1 \mod 8$...
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Brut3Forc3
1948 posts
#5 • 2 Y
Y by Adventure10, Mango247
Oh. I think I misread it and thought that your sets repeated (the whole pattern, not just the plus minus 3 only), but now I get what you mean.
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tc1729
1221 posts
#6 • 1 Y
Y by Adventure10
Solving the first recurrence yields \[x_n=X(-1)^n+Y2^n.\] Using $x_1$ and $x_2$ yields \[x_n=\frac{2^n+(-1)^{n+1}}{3}.\] Similarly, solving for the second recurrence yields \[y_n=2\cdot 3^{n-1}+(-1)^n.\] So if $x_m=y_n$ then $2\cdot 3^n+3(-1)^n=2^m+(-1)^{m+1}$ or $2\cdot 3^n+3(-1)^n+(-1)^m$.

If $m=1$ or $2$, then $n=1$ is the only solution, corresponding the the fact that the term $1$ is in both sequences. If $m>2$, then $2^m\equiv 0\pmod 8$. But we have $3^n\equiv (-1)^n\pmod 4$, so \[2\cdot 3^n+3(-1)^n+(-1)^m\equiv 5(-1)^n+(-1)^m\pmod 8\] which cannot be $0\mod 8$. Hence there are no solutions for $m>2$, and the only integer in both sequences is indeed $1$. $\Box$
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OlympusHero
17020 posts
#7 • 1 Y
Y by Mango247
This is probably overkill, but I still like it :)

Solution
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jred
290 posts
#8
Y by
OlympusHero wrote:
This is probably overkill, but I still like it :)

Solution

But you just proved that $X_n$ cannot be equal to $Y_n$. :blush:
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wateringanddrowned
68 posts
#9 • 1 Y
Y by Mango247
I really want to know how to proof a more general case,what if characteristic roots are irrational?(in this case,someone told me if the sequence is different,they had only finite terms which occur in both sequences.)
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lifeismathematics
1188 posts
#10
Y by
cute sequences
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huashiliao2020
1292 posts
#11
Y by
jred wrote:
OlympusHero wrote:
This is probably overkill, but I still like it :)

Solution

But you just proved that $X_n$ cannot be equal to $Y_n$. :blush:

Yes, that is what we are trying to prove.

Back to the problem. We can prove this by induction. Base Case: Y_1>X_1, Y_2=2x7+3x1=17>3=1+2x1=X_2. Now suppose this is true for some k. Since Y_k>X_k, 2Y_k>X_k, and 3Y_(k-1)>2X_(k-1), so adding these two gives Y_(k+1)>X_(k+1), which means Y_n will always be greater than X_n for all $n\geq1$.

Also, can someone tell me the latex for putting a black/white square at the end of your proofs?

Thanks @below and @2below.
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cinnamon_e
703 posts
#12 • 1 Y
Y by huashiliao2020
white square is \square and black square is \blacksquare :)
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S.Das93
709 posts
#13 • 1 Y
Y by huashiliao2020
huashiliao2020 wrote:

Yes, that is what we are trying to prove.

Back to the problem. We can prove this by induction. Base Case: $Y_1>X_1, Y_2=2\cdot7+3\cdot1=17>3=1+2\cdot1=X_2$. Now suppose this is true for some $k$. Since $Y_k>X_k, 2Y_k>X_k$, and $3Y_{k-1}>2X_{k-1}$, so adding these two gives $Y_{k+1}>X_{k+1}$, which means $Y_n$ will always be greater than $X_n$ for all $n\geq1$. $\blacksquare$

Also, can someone tell me the latex for putting a black/white square at the end of your proofs?

:)
This post has been edited 3 times. Last edited by S.Das93, Apr 13, 2023, 2:43 AM
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