Let $a<b<c$, where $a$ is a positive integer and $b$ and $c$ are odd primes. Prove that if $a mid...
$begingroup$
The prove I tried is the following. I really wish someone can check if I made some logical mistake, especially the last part I found myself diffident proving $a$ can only be $1$ or $5$.
Because $bneq c$ and $b<c$, otherwise switch the value of $b,c$. Becuase $amid (3b+2c)$ and $a mid (3c+2b), exists q_1,q_2 in N$ that $$aq_1=(3b+2c), aq_2=(3c+2b), a(q_2-q_1)=c-b.$$
Therefore we have $amid (c-b)$ and $a mid 3(c-b)+(3b+2c)$ which is $a mid 5c$. And we have $a mid 2(c-b)+(3c+2b)$, $amid 5c$. Hence $a mid 5b$ and $a mid 5c$. Because $b,c$ are distinct odd primes. This is possible only when $amid b$ and $a mid c$ or $a mid 5$. The only number that divides two primes is $1$. And the two numbers that divides $5$ are $1$ and $5$. Hence the two possible values for $a$ are $1$ and $5$.
elementary-number-theory proof-verification
edited Dec 7 '18 at 6:07
Shaun
8,941113682
asked Dec 7 '18 at 5:54
Yupeng WuYupeng Wu
261
$endgroup$
add a comment |
$begingroup$
The prove I tried is the following. I really wish someone can check if I made some logical mistake, especially the last part I found myself diffident proving $a$ can only be $1$ or $5$.
Because $bneq c$ and $b<c$, otherwise switch the value of $b,c$. Becuase $amid (3b+2c)$ and $a mid (3c+2b), exists q_1,q_2 in N$ that $$aq_1=(3b+2c), aq_2=(3c+2b), a(q_2-q_1)=c-b.$$
Therefore we have $amid (c-b)$ and $a mid 3(c-b)+(3b+2c)$ which is $a mid 5c$. And we have $a mid 2(c-b)+(3c+2b)$, $amid 5c$. Hence $a mid 5b$ and $a mid 5c$. Because $b,c$ are distinct odd primes. This is possible only when $amid b$ and $a mid c$ or $a mid 5$. The only number that divides two primes is $1$. And the two numbers that divides $5$ are $1$ and $5$. Hence the two possible values for $a$ are $1$ and $5$.
elementary-number-theory proof-verification
edited Dec 7 '18 at 6:07
Shaun
8,941113682
asked Dec 7 '18 at 5:54
Yupeng WuYupeng Wu
261
$endgroup$
$begingroup$
Your punctuation is a bit off but, other than that, this looks okay. Well done.
$endgroup$
– Shaun
Dec 7 '18 at 6:03
$begingroup$
@Tianlalu, this is elementary number theory. Thenumber-theorytag is mostly redundant once theelementary-number-theorytag is present; besides, the former is more concerned, for instance, with techniques from analysis.
$endgroup$
– Shaun
Dec 7 '18 at 6:07
$begingroup$
@Shaun , got it.
$endgroup$
– Tianlalu
Dec 7 '18 at 6:08
add a comment |
$begingroup$
The prove I tried is the following. I really wish someone can check if I made some logical mistake, especially the last part I found myself diffident proving $a$ can only be $1$ or $5$.
Because $bneq c$ and $b<c$, otherwise switch the value of $b,c$. Becuase $amid (3b+2c)$ and $a mid (3c+2b), exists q_1,q_2 in N$ that $$aq_1=(3b+2c), aq_2=(3c+2b), a(q_2-q_1)=c-b.$$
Therefore we have $amid (c-b)$ and $a mid 3(c-b)+(3b+2c)$ which is $a mid 5c$. And we have $a mid 2(c-b)+(3c+2b)$, $amid 5c$. Hence $a mid 5b$ and $a mid 5c$. Because $b,c$ are distinct odd primes. This is possible only when $amid b$ and $a mid c$ or $a mid 5$. The only number that divides two primes is $1$. And the two numbers that divides $5$ are $1$ and $5$. Hence the two possible values for $a$ are $1$ and $5$.
elementary-number-theory proof-verification
edited Dec 7 '18 at 6:07
Shaun
8,941113682
asked Dec 7 '18 at 5:54
Yupeng WuYupeng Wu
261
$endgroup$
The prove I tried is the following. I really wish someone can check if I made some logical mistake, especially the last part I found myself diffident proving $a$ can only be $1$ or $5$.
Because $bneq c$ and $b<c$, otherwise switch the value of $b,c$. Becuase $amid (3b+2c)$ and $a mid (3c+2b), exists q_1,q_2 in N$ that $$aq_1=(3b+2c), aq_2=(3c+2b), a(q_2-q_1)=c-b.$$
Therefore we have $amid (c-b)$ and $a mid 3(c-b)+(3b+2c)$ which is $a mid 5c$. And we have $a mid 2(c-b)+(3c+2b)$, $amid 5c$. Hence $a mid 5b$ and $a mid 5c$. Because $b,c$ are distinct odd primes. This is possible only when $amid b$ and $a mid c$ or $a mid 5$. The only number that divides two primes is $1$. And the two numbers that divides $5$ are $1$ and $5$. Hence the two possible values for $a$ are $1$ and $5$.
elementary-number-theory proof-verification
elementary-number-theory proof-verification
edited Dec 7 '18 at 6:07
Shaun
8,941113682
asked Dec 7 '18 at 5:54
Yupeng WuYupeng Wu
261
edited Dec 7 '18 at 6:07
Shaun
8,941113682
asked Dec 7 '18 at 5:54
Yupeng WuYupeng Wu
261
edited Dec 7 '18 at 6:07
Shaun
8,941113682
edited Dec 7 '18 at 6:07
Shaun
8,941113682
edited Dec 7 '18 at 6:07
Shaun
8,941113682
8,941113682
asked Dec 7 '18 at 5:54
Yupeng WuYupeng Wu
261
asked Dec 7 '18 at 5:54
Yupeng WuYupeng Wu
261
asked Dec 7 '18 at 5:54
Yupeng WuYupeng Wu
261
261
$begingroup$
Your punctuation is a bit off but, other than that, this looks okay. Well done.
$endgroup$
– Shaun
Dec 7 '18 at 6:03
$begingroup$
@Tianlalu, this is elementary number theory. Thenumber-theorytag is mostly redundant once theelementary-number-theorytag is present; besides, the former is more concerned, for instance, with techniques from analysis.
$endgroup$
– Shaun
Dec 7 '18 at 6:07
$begingroup$
@Shaun , got it.
$endgroup$
– Tianlalu
Dec 7 '18 at 6:08
add a comment |
$begingroup$
Your punctuation is a bit off but, other than that, this looks okay. Well done.
$endgroup$
– Shaun
Dec 7 '18 at 6:03
$begingroup$
@Tianlalu, this is elementary number theory. Thenumber-theorytag is mostly redundant once theelementary-number-theorytag is present; besides, the former is more concerned, for instance, with techniques from analysis.
$endgroup$
– Shaun
Dec 7 '18 at 6:07
$begingroup$
@Shaun , got it.
$endgroup$
– Tianlalu
Dec 7 '18 at 6:08
$begingroup$
Your punctuation is a bit off but, other than that, this looks okay. Well done.
$endgroup$
– Shaun
Dec 7 '18 at 6:03
$begingroup$
Your punctuation is a bit off but, other than that, this looks okay. Well done.
$endgroup$
– Shaun
Dec 7 '18 at 6:03
$begingroup$
@Tianlalu, this is elementary number theory. The
number-theory tag is mostly redundant once the elementary-number-theory tag is present; besides, the former is more concerned, for instance, with techniques from analysis.$endgroup$
– Shaun
Dec 7 '18 at 6:07
$begingroup$
@Tianlalu, this is elementary number theory. The
number-theory tag is mostly redundant once the elementary-number-theory tag is present; besides, the former is more concerned, for instance, with techniques from analysis.$endgroup$
– Shaun
Dec 7 '18 at 6:07
$begingroup$
@Shaun , got it.
$endgroup$
– Tianlalu
Dec 7 '18 at 6:08
$begingroup$
@Shaun , got it.
$endgroup$
– Tianlalu
Dec 7 '18 at 6:08
add a comment |
1 Answer
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$begingroup$
You proof is fine, but you are over elaborating at certain points. Do you need $q_1,q_2$, for example? You do not seem to have used them.
Indeed, the simplest proof is that if $a|3b+2c$ and $a | 3c+2b$, then $a | 3(3c+2b) - 2(3b+2c) = 5c$, and $a | 3(3b+2c) - 2(3c + 2b) = 5b$.
Therefore, $a | 5c$ and $a | 5b$, which means that $a$ divides the greatest common divisor of $5c$ and $5b$. But, $(5c,5b) = 5(c,b) = 5 times 1 = 5$ (as $c,b$ are prime), so $a | 5$.
You have done the same thing, but with a little more elaboration.
Also, it is useful to show that both cases are attained : for example, $b = 2,c=7$ gives $3b+2c = 20$, and $3c+2b = 25$.
However, note that if $5 | 3c+2b$ and $5 | 3b+2c$, then $5 | b-c$. Given that $b,c$ are primes, this forces one of $b,c$ to be even. Thus, $b = 2$ and $c = 7$ is forced.
Conclusion : if $b neq 2, c neq 7$, then in fact $a = 1$ is forced given $b < c$.
Try this more general question : given $ b < c$ natural numbers not necessarily prime, and integers $d,e$ , if $a | db+ce$ and $a | eb+cd$, what can you say about $a$?
answered Dec 7 '18 at 7:18
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.9k33376
$endgroup$
add a comment |
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1 Answer
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$begingroup$
You proof is fine, but you are over elaborating at certain points. Do you need $q_1,q_2$, for example? You do not seem to have used them.
Indeed, the simplest proof is that if $a|3b+2c$ and $a | 3c+2b$, then $a | 3(3c+2b) - 2(3b+2c) = 5c$, and $a | 3(3b+2c) - 2(3c + 2b) = 5b$.
Therefore, $a | 5c$ and $a | 5b$, which means that $a$ divides the greatest common divisor of $5c$ and $5b$. But, $(5c,5b) = 5(c,b) = 5 times 1 = 5$ (as $c,b$ are prime), so $a | 5$.
You have done the same thing, but with a little more elaboration.
Also, it is useful to show that both cases are attained : for example, $b = 2,c=7$ gives $3b+2c = 20$, and $3c+2b = 25$.
However, note that if $5 | 3c+2b$ and $5 | 3b+2c$, then $5 | b-c$. Given that $b,c$ are primes, this forces one of $b,c$ to be even. Thus, $b = 2$ and $c = 7$ is forced.
Conclusion : if $b neq 2, c neq 7$, then in fact $a = 1$ is forced given $b < c$.
Try this more general question : given $ b < c$ natural numbers not necessarily prime, and integers $d,e$ , if $a | db+ce$ and $a | eb+cd$, what can you say about $a$?
answered Dec 7 '18 at 7:18
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.9k33376
$endgroup$
add a comment |
$begingroup$
You proof is fine, but you are over elaborating at certain points. Do you need $q_1,q_2$, for example? You do not seem to have used them.
Indeed, the simplest proof is that if $a|3b+2c$ and $a | 3c+2b$, then $a | 3(3c+2b) - 2(3b+2c) = 5c$, and $a | 3(3b+2c) - 2(3c + 2b) = 5b$.
Therefore, $a | 5c$ and $a | 5b$, which means that $a$ divides the greatest common divisor of $5c$ and $5b$. But, $(5c,5b) = 5(c,b) = 5 times 1 = 5$ (as $c,b$ are prime), so $a | 5$.
You have done the same thing, but with a little more elaboration.
Also, it is useful to show that both cases are attained : for example, $b = 2,c=7$ gives $3b+2c = 20$, and $3c+2b = 25$.
However, note that if $5 | 3c+2b$ and $5 | 3b+2c$, then $5 | b-c$. Given that $b,c$ are primes, this forces one of $b,c$ to be even. Thus, $b = 2$ and $c = 7$ is forced.
Conclusion : if $b neq 2, c neq 7$, then in fact $a = 1$ is forced given $b < c$.
Try this more general question : given $ b < c$ natural numbers not necessarily prime, and integers $d,e$ , if $a | db+ce$ and $a | eb+cd$, what can you say about $a$?
answered Dec 7 '18 at 7:18
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.9k33376
$endgroup$
add a comment |
$begingroup$
You proof is fine, but you are over elaborating at certain points. Do you need $q_1,q_2$, for example? You do not seem to have used them.
Indeed, the simplest proof is that if $a|3b+2c$ and $a | 3c+2b$, then $a | 3(3c+2b) - 2(3b+2c) = 5c$, and $a | 3(3b+2c) - 2(3c + 2b) = 5b$.
Therefore, $a | 5c$ and $a | 5b$, which means that $a$ divides the greatest common divisor of $5c$ and $5b$. But, $(5c,5b) = 5(c,b) = 5 times 1 = 5$ (as $c,b$ are prime), so $a | 5$.
You have done the same thing, but with a little more elaboration.
Also, it is useful to show that both cases are attained : for example, $b = 2,c=7$ gives $3b+2c = 20$, and $3c+2b = 25$.
However, note that if $5 | 3c+2b$ and $5 | 3b+2c$, then $5 | b-c$. Given that $b,c$ are primes, this forces one of $b,c$ to be even. Thus, $b = 2$ and $c = 7$ is forced.
Conclusion : if $b neq 2, c neq 7$, then in fact $a = 1$ is forced given $b < c$.
Try this more general question : given $ b < c$ natural numbers not necessarily prime, and integers $d,e$ , if $a | db+ce$ and $a | eb+cd$, what can you say about $a$?
answered Dec 7 '18 at 7:18
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.9k33376
$endgroup$
You proof is fine, but you are over elaborating at certain points. Do you need $q_1,q_2$, for example? You do not seem to have used them.
Indeed, the simplest proof is that if $a|3b+2c$ and $a | 3c+2b$, then $a | 3(3c+2b) - 2(3b+2c) = 5c$, and $a | 3(3b+2c) - 2(3c + 2b) = 5b$.
Therefore, $a | 5c$ and $a | 5b$, which means that $a$ divides the greatest common divisor of $5c$ and $5b$. But, $(5c,5b) = 5(c,b) = 5 times 1 = 5$ (as $c,b$ are prime), so $a | 5$.
You have done the same thing, but with a little more elaboration.
Also, it is useful to show that both cases are attained : for example, $b = 2,c=7$ gives $3b+2c = 20$, and $3c+2b = 25$.
However, note that if $5 | 3c+2b$ and $5 | 3b+2c$, then $5 | b-c$. Given that $b,c$ are primes, this forces one of $b,c$ to be even. Thus, $b = 2$ and $c = 7$ is forced.
Conclusion : if $b neq 2, c neq 7$, then in fact $a = 1$ is forced given $b < c$.
Try this more general question : given $ b < c$ natural numbers not necessarily prime, and integers $d,e$ , if $a | db+ce$ and $a | eb+cd$, what can you say about $a$?
answered Dec 7 '18 at 7:18
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.9k33376
answered Dec 7 '18 at 7:18
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.9k33376
answered Dec 7 '18 at 7:18
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.9k33376
answered Dec 7 '18 at 7:18
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.9k33376
37.9k33376
add a comment |
add a comment |
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$begingroup$
Your punctuation is a bit off but, other than that, this looks okay. Well done.
$endgroup$
– Shaun
Dec 7 '18 at 6:03
$begingroup$
@Tianlalu, this is elementary number theory. The
number-theorytag is mostly redundant once theelementary-number-theorytag is present; besides, the former is more concerned, for instance, with techniques from analysis.$endgroup$
– Shaun
Dec 7 '18 at 6:07
$begingroup$
@Shaun , got it.
$endgroup$
– Tianlalu
Dec 7 '18 at 6:08