show this either $m=k^2$ or $m = 5k^2$ for some integer $k$.Is this proof of $a^1/2$ being either integer or irrational circular/incorrect?How many integer solutions of “a” exist for this equation?Show that for any positive integer n, $(3n)!/(3!)^n$ is an integer.Prove that if $a_0geq 2$, $f(n)$ is not prime for some integer $z$.Show that if $a$ is a positive integer and $a^m + 1$ is an odd prime, then $m = 2^n$ for some non-negative integer $n$Is there a method to find all primitive roots for some integer $n$?Show that only one prime can be expressed as $n^3-1$ for some positive integer $n$Show that $(sqrt2+1)^n = sqrtm+sqrtm-1$ for some positive integer $m$Find the largest possible integer n such that $sqrtn+sqrtn+60=m$ for some non-square integer m.Show that $omega * mu$ is either 0 or 1.
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show this either $m=k^2$ or $m = 5k^2$ for some integer $k$.
Is this proof of $a^1/2$ being either integer or irrational circular/incorrect?How many integer solutions of “a” exist for this equation?Show that for any positive integer n, $(3n)!/(3!)^n$ is an integer.Prove that if $a_0geq 2$, $f(n)$ is not prime for some integer $z$.Show that if $a$ is a positive integer and $a^m + 1$ is an odd prime, then $m = 2^n$ for some non-negative integer $n$Is there a method to find all primitive roots for some integer $n$?Show that only one prime can be expressed as $n^3-1$ for some positive integer $n$Show that $(sqrt2+1)^n = sqrtm+sqrtm-1$ for some positive integer $m$Find the largest possible integer n such that $sqrtn+sqrtn+60=m$ for some non-square integer m.Show that $omega * mu$ is either 0 or 1.
$begingroup$
Let $f(x)$ be the distance from $x$ to the nearest perfect square. For example, $f(pi) = 4 - pi$. Let $alpha = frac3 + sqrt52$ and let $m$ be an integer such that the sequence $a_n = f(m ; alpha^n)$ is bounded. Prove that either $m=k^2$ or $m = 5k^2$ for some integer $k$.
This problem from:
The 58th IMO in 2017 was loaded into IMO history with its high difficulty and low score. The IMO set a number of records since the restructuring in 1983, including the lowest total score(170 points), the lowest personal score(35 points), and the lowest gold score(25 points). Only 2 of the most difficult 3 questions scored full marks. The total score of 615 players in 111 participating countries worldwide was only 26 points, with an average score of 0.042 points. The top 10 countries scored a total of 6 points on this question.
After the match, a Brazilian player Dave Sena, who won a silver medal with 19 points, proposed the "IMO Revenge" to express his respect for the teachers and his "revenge". He collected questions from IMO players and selected four questions as trial questions. Invite and encourage team leaders, deputy team leaders and observers to participate. As a result, the initiative received a lot of positive response from IMO players and team leaders.
number-theory
$endgroup$
add a comment |
$begingroup$
Let $f(x)$ be the distance from $x$ to the nearest perfect square. For example, $f(pi) = 4 - pi$. Let $alpha = frac3 + sqrt52$ and let $m$ be an integer such that the sequence $a_n = f(m ; alpha^n)$ is bounded. Prove that either $m=k^2$ or $m = 5k^2$ for some integer $k$.
This problem from:
The 58th IMO in 2017 was loaded into IMO history with its high difficulty and low score. The IMO set a number of records since the restructuring in 1983, including the lowest total score(170 points), the lowest personal score(35 points), and the lowest gold score(25 points). Only 2 of the most difficult 3 questions scored full marks. The total score of 615 players in 111 participating countries worldwide was only 26 points, with an average score of 0.042 points. The top 10 countries scored a total of 6 points on this question.
After the match, a Brazilian player Dave Sena, who won a silver medal with 19 points, proposed the "IMO Revenge" to express his respect for the teachers and his "revenge". He collected questions from IMO players and selected four questions as trial questions. Invite and encourage team leaders, deputy team leaders and observers to participate. As a result, the initiative received a lot of positive response from IMO players and team leaders.
number-theory
$endgroup$
1
$begingroup$
I don't get it, there is no such problem in IMO 2017, can you add some reference? Also AOPS forum (artofproblemsolving.com/community) might be better platform for such problems, MSE is more about asking math questions, less about just solving stuff (although it often ends that way). Nevermind, found it (on AOPS forum): artofproblemsolving.com/community/q3h1482029p8654340. There is a link to pdf at the bottom. And here is this specific problem: artofproblemsolving.com/community/q4h1482444p8659066 .
$endgroup$
– Sil
Mar 17 at 16:33
add a comment |
$begingroup$
Let $f(x)$ be the distance from $x$ to the nearest perfect square. For example, $f(pi) = 4 - pi$. Let $alpha = frac3 + sqrt52$ and let $m$ be an integer such that the sequence $a_n = f(m ; alpha^n)$ is bounded. Prove that either $m=k^2$ or $m = 5k^2$ for some integer $k$.
This problem from:
The 58th IMO in 2017 was loaded into IMO history with its high difficulty and low score. The IMO set a number of records since the restructuring in 1983, including the lowest total score(170 points), the lowest personal score(35 points), and the lowest gold score(25 points). Only 2 of the most difficult 3 questions scored full marks. The total score of 615 players in 111 participating countries worldwide was only 26 points, with an average score of 0.042 points. The top 10 countries scored a total of 6 points on this question.
After the match, a Brazilian player Dave Sena, who won a silver medal with 19 points, proposed the "IMO Revenge" to express his respect for the teachers and his "revenge". He collected questions from IMO players and selected four questions as trial questions. Invite and encourage team leaders, deputy team leaders and observers to participate. As a result, the initiative received a lot of positive response from IMO players and team leaders.
number-theory
$endgroup$
Let $f(x)$ be the distance from $x$ to the nearest perfect square. For example, $f(pi) = 4 - pi$. Let $alpha = frac3 + sqrt52$ and let $m$ be an integer such that the sequence $a_n = f(m ; alpha^n)$ is bounded. Prove that either $m=k^2$ or $m = 5k^2$ for some integer $k$.
This problem from:
The 58th IMO in 2017 was loaded into IMO history with its high difficulty and low score. The IMO set a number of records since the restructuring in 1983, including the lowest total score(170 points), the lowest personal score(35 points), and the lowest gold score(25 points). Only 2 of the most difficult 3 questions scored full marks. The total score of 615 players in 111 participating countries worldwide was only 26 points, with an average score of 0.042 points. The top 10 countries scored a total of 6 points on this question.
After the match, a Brazilian player Dave Sena, who won a silver medal with 19 points, proposed the "IMO Revenge" to express his respect for the teachers and his "revenge". He collected questions from IMO players and selected four questions as trial questions. Invite and encourage team leaders, deputy team leaders and observers to participate. As a result, the initiative received a lot of positive response from IMO players and team leaders.
number-theory
number-theory
asked Mar 17 at 14:01
inequalityinequality
790522
790522
1
$begingroup$
I don't get it, there is no such problem in IMO 2017, can you add some reference? Also AOPS forum (artofproblemsolving.com/community) might be better platform for such problems, MSE is more about asking math questions, less about just solving stuff (although it often ends that way). Nevermind, found it (on AOPS forum): artofproblemsolving.com/community/q3h1482029p8654340. There is a link to pdf at the bottom. And here is this specific problem: artofproblemsolving.com/community/q4h1482444p8659066 .
$endgroup$
– Sil
Mar 17 at 16:33
add a comment |
1
$begingroup$
I don't get it, there is no such problem in IMO 2017, can you add some reference? Also AOPS forum (artofproblemsolving.com/community) might be better platform for such problems, MSE is more about asking math questions, less about just solving stuff (although it often ends that way). Nevermind, found it (on AOPS forum): artofproblemsolving.com/community/q3h1482029p8654340. There is a link to pdf at the bottom. And here is this specific problem: artofproblemsolving.com/community/q4h1482444p8659066 .
$endgroup$
– Sil
Mar 17 at 16:33
1
1
$begingroup$
I don't get it, there is no such problem in IMO 2017, can you add some reference? Also AOPS forum (artofproblemsolving.com/community) might be better platform for such problems, MSE is more about asking math questions, less about just solving stuff (although it often ends that way). Nevermind, found it (on AOPS forum): artofproblemsolving.com/community/q3h1482029p8654340. There is a link to pdf at the bottom. And here is this specific problem: artofproblemsolving.com/community/q4h1482444p8659066 .
$endgroup$
– Sil
Mar 17 at 16:33
$begingroup$
I don't get it, there is no such problem in IMO 2017, can you add some reference? Also AOPS forum (artofproblemsolving.com/community) might be better platform for such problems, MSE is more about asking math questions, less about just solving stuff (although it often ends that way). Nevermind, found it (on AOPS forum): artofproblemsolving.com/community/q3h1482029p8654340. There is a link to pdf at the bottom. And here is this specific problem: artofproblemsolving.com/community/q4h1482444p8659066 .
$endgroup$
– Sil
Mar 17 at 16:33
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Solutions to this problem are provided in the commented link. This answer is to prove little bit more in the context of Pisot-Vijayaraghavan number.
The function $f(x)$ can be written as
$$
f(malpha^n)=min_zinmathbbZ|(z-sqrt m beta^n)(z+sqrt mbeta^n)|$$
$$=begincases |sqrt m beta^n | 2sqrt m beta^n + O(1) &mbox in all cases, \
| sqrt m beta^n|(epsilon_m,n|sqrt m beta^n|+2sqrt m beta^n) &mboxsqrt m beta^n endcases.
$$
where $beta=frac1+sqrt 52$ and $overlinebeta=frac1-sqrt 52$, $|x|$ is the distance from $x$ to the nearest integer to $x$. Here, $epsilon_m,n=pm 1$. The implied constant in the big-Oh is at most $1/4$.
For $f(malpha^n)$ to be bounded, we must have $|sqrt m beta^n|rightarrow 0$ as $nrightarrow infty$. This is because $beta^nrightarrowinfty$, $overlinebeta^nrightarrow 0$.
By one of the theorem that wiki link has, we must have $sqrt m in mathbbQ(sqrt 5)$. This holds only when $m=k^2$ or $m=5k^2$.
Conversely, we assume $m=k^2$ or $m=5k^2$. Then we have
$$
|sqrt m beta^n|=sqrt m |overlinebeta^n|
$$
since $sqrt m beta^n+sqrt m overlinebeta^n in mathbbZ$ if $m=k^2$, and $sqrt m beta^n-sqrt m overlinebeta^ninmathbbZ$ if $m=5k^2$.
Now, by $betaoverlinebeta=-1$,
$$
f(malpha^n)=| sqrt m beta^n|(epsilon_m,n|sqrt m beta^n|+2sqrt m beta^n)rightarrow 2m textrmas nrightarrowinfty.
$$
$endgroup$
add a comment |
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$begingroup$
Solutions to this problem are provided in the commented link. This answer is to prove little bit more in the context of Pisot-Vijayaraghavan number.
The function $f(x)$ can be written as
$$
f(malpha^n)=min_zinmathbbZ|(z-sqrt m beta^n)(z+sqrt mbeta^n)|$$
$$=begincases |sqrt m beta^n | 2sqrt m beta^n + O(1) &mbox in all cases, \
| sqrt m beta^n|(epsilon_m,n|sqrt m beta^n|+2sqrt m beta^n) &mboxsqrt m beta^n endcases.
$$
where $beta=frac1+sqrt 52$ and $overlinebeta=frac1-sqrt 52$, $|x|$ is the distance from $x$ to the nearest integer to $x$. Here, $epsilon_m,n=pm 1$. The implied constant in the big-Oh is at most $1/4$.
For $f(malpha^n)$ to be bounded, we must have $|sqrt m beta^n|rightarrow 0$ as $nrightarrow infty$. This is because $beta^nrightarrowinfty$, $overlinebeta^nrightarrow 0$.
By one of the theorem that wiki link has, we must have $sqrt m in mathbbQ(sqrt 5)$. This holds only when $m=k^2$ or $m=5k^2$.
Conversely, we assume $m=k^2$ or $m=5k^2$. Then we have
$$
|sqrt m beta^n|=sqrt m |overlinebeta^n|
$$
since $sqrt m beta^n+sqrt m overlinebeta^n in mathbbZ$ if $m=k^2$, and $sqrt m beta^n-sqrt m overlinebeta^ninmathbbZ$ if $m=5k^2$.
Now, by $betaoverlinebeta=-1$,
$$
f(malpha^n)=| sqrt m beta^n|(epsilon_m,n|sqrt m beta^n|+2sqrt m beta^n)rightarrow 2m textrmas nrightarrowinfty.
$$
$endgroup$
add a comment |
$begingroup$
Solutions to this problem are provided in the commented link. This answer is to prove little bit more in the context of Pisot-Vijayaraghavan number.
The function $f(x)$ can be written as
$$
f(malpha^n)=min_zinmathbbZ|(z-sqrt m beta^n)(z+sqrt mbeta^n)|$$
$$=begincases |sqrt m beta^n | 2sqrt m beta^n + O(1) &mbox in all cases, \
| sqrt m beta^n|(epsilon_m,n|sqrt m beta^n|+2sqrt m beta^n) &mboxsqrt m beta^n endcases.
$$
where $beta=frac1+sqrt 52$ and $overlinebeta=frac1-sqrt 52$, $|x|$ is the distance from $x$ to the nearest integer to $x$. Here, $epsilon_m,n=pm 1$. The implied constant in the big-Oh is at most $1/4$.
For $f(malpha^n)$ to be bounded, we must have $|sqrt m beta^n|rightarrow 0$ as $nrightarrow infty$. This is because $beta^nrightarrowinfty$, $overlinebeta^nrightarrow 0$.
By one of the theorem that wiki link has, we must have $sqrt m in mathbbQ(sqrt 5)$. This holds only when $m=k^2$ or $m=5k^2$.
Conversely, we assume $m=k^2$ or $m=5k^2$. Then we have
$$
|sqrt m beta^n|=sqrt m |overlinebeta^n|
$$
since $sqrt m beta^n+sqrt m overlinebeta^n in mathbbZ$ if $m=k^2$, and $sqrt m beta^n-sqrt m overlinebeta^ninmathbbZ$ if $m=5k^2$.
Now, by $betaoverlinebeta=-1$,
$$
f(malpha^n)=| sqrt m beta^n|(epsilon_m,n|sqrt m beta^n|+2sqrt m beta^n)rightarrow 2m textrmas nrightarrowinfty.
$$
$endgroup$
add a comment |
$begingroup$
Solutions to this problem are provided in the commented link. This answer is to prove little bit more in the context of Pisot-Vijayaraghavan number.
The function $f(x)$ can be written as
$$
f(malpha^n)=min_zinmathbbZ|(z-sqrt m beta^n)(z+sqrt mbeta^n)|$$
$$=begincases |sqrt m beta^n | 2sqrt m beta^n + O(1) &mbox in all cases, \
| sqrt m beta^n|(epsilon_m,n|sqrt m beta^n|+2sqrt m beta^n) &mboxsqrt m beta^n endcases.
$$
where $beta=frac1+sqrt 52$ and $overlinebeta=frac1-sqrt 52$, $|x|$ is the distance from $x$ to the nearest integer to $x$. Here, $epsilon_m,n=pm 1$. The implied constant in the big-Oh is at most $1/4$.
For $f(malpha^n)$ to be bounded, we must have $|sqrt m beta^n|rightarrow 0$ as $nrightarrow infty$. This is because $beta^nrightarrowinfty$, $overlinebeta^nrightarrow 0$.
By one of the theorem that wiki link has, we must have $sqrt m in mathbbQ(sqrt 5)$. This holds only when $m=k^2$ or $m=5k^2$.
Conversely, we assume $m=k^2$ or $m=5k^2$. Then we have
$$
|sqrt m beta^n|=sqrt m |overlinebeta^n|
$$
since $sqrt m beta^n+sqrt m overlinebeta^n in mathbbZ$ if $m=k^2$, and $sqrt m beta^n-sqrt m overlinebeta^ninmathbbZ$ if $m=5k^2$.
Now, by $betaoverlinebeta=-1$,
$$
f(malpha^n)=| sqrt m beta^n|(epsilon_m,n|sqrt m beta^n|+2sqrt m beta^n)rightarrow 2m textrmas nrightarrowinfty.
$$
$endgroup$
Solutions to this problem are provided in the commented link. This answer is to prove little bit more in the context of Pisot-Vijayaraghavan number.
The function $f(x)$ can be written as
$$
f(malpha^n)=min_zinmathbbZ|(z-sqrt m beta^n)(z+sqrt mbeta^n)|$$
$$=begincases |sqrt m beta^n | 2sqrt m beta^n + O(1) &mbox in all cases, \
| sqrt m beta^n|(epsilon_m,n|sqrt m beta^n|+2sqrt m beta^n) &mboxsqrt m beta^n endcases.
$$
where $beta=frac1+sqrt 52$ and $overlinebeta=frac1-sqrt 52$, $|x|$ is the distance from $x$ to the nearest integer to $x$. Here, $epsilon_m,n=pm 1$. The implied constant in the big-Oh is at most $1/4$.
For $f(malpha^n)$ to be bounded, we must have $|sqrt m beta^n|rightarrow 0$ as $nrightarrow infty$. This is because $beta^nrightarrowinfty$, $overlinebeta^nrightarrow 0$.
By one of the theorem that wiki link has, we must have $sqrt m in mathbbQ(sqrt 5)$. This holds only when $m=k^2$ or $m=5k^2$.
Conversely, we assume $m=k^2$ or $m=5k^2$. Then we have
$$
|sqrt m beta^n|=sqrt m |overlinebeta^n|
$$
since $sqrt m beta^n+sqrt m overlinebeta^n in mathbbZ$ if $m=k^2$, and $sqrt m beta^n-sqrt m overlinebeta^ninmathbbZ$ if $m=5k^2$.
Now, by $betaoverlinebeta=-1$,
$$
f(malpha^n)=| sqrt m beta^n|(epsilon_m,n|sqrt m beta^n|+2sqrt m beta^n)rightarrow 2m textrmas nrightarrowinfty.
$$
edited Mar 18 at 21:58
answered Mar 18 at 6:53
i707107i707107
12.5k21647
12.5k21647
add a comment |
add a comment |
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$begingroup$
I don't get it, there is no such problem in IMO 2017, can you add some reference? Also AOPS forum (artofproblemsolving.com/community) might be better platform for such problems, MSE is more about asking math questions, less about just solving stuff (although it often ends that way). Nevermind, found it (on AOPS forum): artofproblemsolving.com/community/q3h1482029p8654340. There is a link to pdf at the bottom. And here is this specific problem: artofproblemsolving.com/community/q4h1482444p8659066 .
$endgroup$
– Sil
Mar 17 at 16:33