Given prime numbers $p,q$ and $r$, $p|qr − 1$, $q|rp − 1$, and $r|pq − 1$. What is the value of all possible $pqr$?What is the value of $Sigma frac1pqr$ with $frac1a>frac1pqr$?Can one show a beginning student how to use the $p$-adics to solve a problem?Evaluate $lim_n to inftyBigl[bigl(prod_k=1^nfrac2k2k-1bigr)bigl(int_-1^inftyfrac(cosx)^2n2^x,dxbigr)Bigr]$Mathematical Reflections J339: Solving $fracx-1y+1 + fracy-1z+1 + fracz-1x+1 = 1$ in positive integersProof verification - working with integersFind all pair of $(x,y)$ such that $fracx^2+y^2x-y$ is an integer and divides 1995.Find all pairs of prime numbers $p$ and $q$ such that $,p^2-p-1=q^3.$Simple solution to Question 6 from the 1988 Math OlympiadNumber Theory : Prime Divisor Property ProofShowing that $a^2+b^2+c^2+d^2+e^2+65=abcde$ has integer solutions greater than $2018$?
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Given prime numbers $p,q$ and $r$, $p|qr − 1$, $q|rp − 1$, and $r|pq − 1$. What is the value of all possible $pqr$?
What is the value of $Sigma frac1pqr$ with $frac1a>frac1pqr$?Can one show a beginning student how to use the $p$-adics to solve a problem?Evaluate $lim_n to inftyBigl[bigl(prod_k=1^nfrac2k2k-1bigr)bigl(int_-1^inftyfrac(cosx)^2n2^x,dxbigr)Bigr]$Mathematical Reflections J339: Solving $fracx-1y+1 + fracy-1z+1 + fracz-1x+1 = 1$ in positive integersProof verification - working with integersFind all pair of $(x,y)$ such that $fracx^2+y^2x-y$ is an integer and divides 1995.Find all pairs of prime numbers $p$ and $q$ such that $,p^2-p-1=q^3.$Simple solution to Question 6 from the 1988 Math OlympiadNumber Theory : Prime Divisor Property ProofShowing that $a^2+b^2+c^2+d^2+e^2+65=abcde$ has integer solutions greater than $2018$?
$begingroup$
Let $p,q$ and $r$ be prime numbers. It is given that $p$ divides $qr − 1$, $q$ divides $rp − 1$, and $r$ divides $pq − 1$.
Determine all possible values of $pqr$.
I think I'm missing something in this problem. There's no solutions online so I was wondering if you guys could show me what I'm doing wrong.
We must have one of $p,q,r = 2$ as $pq - 1$, $pr - 1$, $qr - 1$ are congruent to $0 mod 2$ if $p,q,r > 2$, so wlog let $p = 2$ (more than one of $p,q,r$ being 2 gives contradiction). Then $q$ divides $2r - 1$ (so $2r - 1 = xq$) and $r$ divides $2q - 1$ (so $2q - 1 = yr$). Solving for r we have that $(xq + 1)/2 = (2q - 1)/y$ hence $ q = (2 + y)/(4 - xy)$. This gives the only pair as $(3,5)$, so $pqr$ can only be $30$?
The way the question is phrased leads me to believe there's more to it, and also it's the last question on an olympiad so it should take longer than 5 minutes.
Re-did it, is this okay?
If we write $ rq - 1 = px $ and $ rp - 1 = qy$ we obtain $ p = fracr + yr^2 - xy $ and $ q = fracr + xr^2 - xy $ (and clearly we can't have $ r^2 = xy $) so $q(r + y) = p(r + x)$. Then subbing $ x = frac3q - 1p $ and $y = frac3p - 1q$ we obtain $r=3$ if $p$ doesn't equal $q$ (if $p=q$ then no solutions). Then it's just a matter of checking cases and we just get $(p,q,r) = (5,7,3), (2,5,3)$. But $(5,7,3)$ doesn't satisfy the fact that $r$ divides $pq - 1$ hence the only solution is $pqr = 30$. I hope that's ok
number-theory contest-math
edited Oct 11 '16 at 6:29
Irregular User
2,88251843
asked Jul 20 '14 at 18:18
user121591user121591
737
$endgroup$
|
show 4 more comments
$begingroup$
Let $p,q$ and $r$ be prime numbers. It is given that $p$ divides $qr − 1$, $q$ divides $rp − 1$, and $r$ divides $pq − 1$.
Determine all possible values of $pqr$.
I think I'm missing something in this problem. There's no solutions online so I was wondering if you guys could show me what I'm doing wrong.
We must have one of $p,q,r = 2$ as $pq - 1$, $pr - 1$, $qr - 1$ are congruent to $0 mod 2$ if $p,q,r > 2$, so wlog let $p = 2$ (more than one of $p,q,r$ being 2 gives contradiction). Then $q$ divides $2r - 1$ (so $2r - 1 = xq$) and $r$ divides $2q - 1$ (so $2q - 1 = yr$). Solving for r we have that $(xq + 1)/2 = (2q - 1)/y$ hence $ q = (2 + y)/(4 - xy)$. This gives the only pair as $(3,5)$, so $pqr$ can only be $30$?
The way the question is phrased leads me to believe there's more to it, and also it's the last question on an olympiad so it should take longer than 5 minutes.
Re-did it, is this okay?
If we write $ rq - 1 = px $ and $ rp - 1 = qy$ we obtain $ p = fracr + yr^2 - xy $ and $ q = fracr + xr^2 - xy $ (and clearly we can't have $ r^2 = xy $) so $q(r + y) = p(r + x)$. Then subbing $ x = frac3q - 1p $ and $y = frac3p - 1q$ we obtain $r=3$ if $p$ doesn't equal $q$ (if $p=q$ then no solutions). Then it's just a matter of checking cases and we just get $(p,q,r) = (5,7,3), (2,5,3)$. But $(5,7,3)$ doesn't satisfy the fact that $r$ divides $pq - 1$ hence the only solution is $pqr = 30$. I hope that's ok
number-theory contest-math
edited Oct 11 '16 at 6:29
Irregular User
2,88251843
asked Jul 20 '14 at 18:18
user121591user121591
737
$endgroup$
$begingroup$
Here's a MathJax tutorial :)
$endgroup$
– Shaun
Jul 20 '14 at 18:21
$begingroup$
Are negative primes allowed?
$endgroup$
– Mathmo123
Jul 20 '14 at 18:23
$begingroup$
Isn't a prime number positive by definition? Thanks Shaun I tried to make it better
$endgroup$
– user121591
Jul 20 '14 at 18:25
1
$begingroup$
I don't think your first line is correct. Suppose they are all greater than two. Then indeed all of the products of the form $pq-1$ are even. But that isn't a problem (even numbers can have odd prime divisors), and it doesn't immediately imply anything interesting.
$endgroup$
– Potato
Jul 20 '14 at 18:35
$begingroup$
By the way, I think you can piece a solution together from the comments here: nrich.maths.org/discus/messages/153904/149403.html?1281107322
$endgroup$
– Potato
Jul 20 '14 at 18:40
|
show 4 more comments
$begingroup$
Let $p,q$ and $r$ be prime numbers. It is given that $p$ divides $qr − 1$, $q$ divides $rp − 1$, and $r$ divides $pq − 1$.
Determine all possible values of $pqr$.
I think I'm missing something in this problem. There's no solutions online so I was wondering if you guys could show me what I'm doing wrong.
We must have one of $p,q,r = 2$ as $pq - 1$, $pr - 1$, $qr - 1$ are congruent to $0 mod 2$ if $p,q,r > 2$, so wlog let $p = 2$ (more than one of $p,q,r$ being 2 gives contradiction). Then $q$ divides $2r - 1$ (so $2r - 1 = xq$) and $r$ divides $2q - 1$ (so $2q - 1 = yr$). Solving for r we have that $(xq + 1)/2 = (2q - 1)/y$ hence $ q = (2 + y)/(4 - xy)$. This gives the only pair as $(3,5)$, so $pqr$ can only be $30$?
The way the question is phrased leads me to believe there's more to it, and also it's the last question on an olympiad so it should take longer than 5 minutes.
Re-did it, is this okay?
If we write $ rq - 1 = px $ and $ rp - 1 = qy$ we obtain $ p = fracr + yr^2 - xy $ and $ q = fracr + xr^2 - xy $ (and clearly we can't have $ r^2 = xy $) so $q(r + y) = p(r + x)$. Then subbing $ x = frac3q - 1p $ and $y = frac3p - 1q$ we obtain $r=3$ if $p$ doesn't equal $q$ (if $p=q$ then no solutions). Then it's just a matter of checking cases and we just get $(p,q,r) = (5,7,3), (2,5,3)$. But $(5,7,3)$ doesn't satisfy the fact that $r$ divides $pq - 1$ hence the only solution is $pqr = 30$. I hope that's ok
number-theory contest-math
edited Oct 11 '16 at 6:29
Irregular User
2,88251843
asked Jul 20 '14 at 18:18
user121591user121591
737
$endgroup$
Let $p,q$ and $r$ be prime numbers. It is given that $p$ divides $qr − 1$, $q$ divides $rp − 1$, and $r$ divides $pq − 1$.
Determine all possible values of $pqr$.
I think I'm missing something in this problem. There's no solutions online so I was wondering if you guys could show me what I'm doing wrong.
We must have one of $p,q,r = 2$ as $pq - 1$, $pr - 1$, $qr - 1$ are congruent to $0 mod 2$ if $p,q,r > 2$, so wlog let $p = 2$ (more than one of $p,q,r$ being 2 gives contradiction). Then $q$ divides $2r - 1$ (so $2r - 1 = xq$) and $r$ divides $2q - 1$ (so $2q - 1 = yr$). Solving for r we have that $(xq + 1)/2 = (2q - 1)/y$ hence $ q = (2 + y)/(4 - xy)$. This gives the only pair as $(3,5)$, so $pqr$ can only be $30$?
The way the question is phrased leads me to believe there's more to it, and also it's the last question on an olympiad so it should take longer than 5 minutes.
Re-did it, is this okay?
If we write $ rq - 1 = px $ and $ rp - 1 = qy$ we obtain $ p = fracr + yr^2 - xy $ and $ q = fracr + xr^2 - xy $ (and clearly we can't have $ r^2 = xy $) so $q(r + y) = p(r + x)$. Then subbing $ x = frac3q - 1p $ and $y = frac3p - 1q$ we obtain $r=3$ if $p$ doesn't equal $q$ (if $p=q$ then no solutions). Then it's just a matter of checking cases and we just get $(p,q,r) = (5,7,3), (2,5,3)$. But $(5,7,3)$ doesn't satisfy the fact that $r$ divides $pq - 1$ hence the only solution is $pqr = 30$. I hope that's ok
number-theory contest-math
number-theory contest-math
edited Oct 11 '16 at 6:29
Irregular User
2,88251843
asked Jul 20 '14 at 18:18
user121591user121591
737
edited Oct 11 '16 at 6:29
Irregular User
2,88251843
asked Jul 20 '14 at 18:18
user121591user121591
737
edited Oct 11 '16 at 6:29
Irregular User
2,88251843
edited Oct 11 '16 at 6:29
Irregular User
2,88251843
edited Oct 11 '16 at 6:29
Irregular User
2,88251843
2,88251843
asked Jul 20 '14 at 18:18
user121591user121591
737
asked Jul 20 '14 at 18:18
user121591user121591
737
asked Jul 20 '14 at 18:18
user121591user121591
737
737
$begingroup$
Here's a MathJax tutorial :)
$endgroup$
– Shaun
Jul 20 '14 at 18:21
$begingroup$
Are negative primes allowed?
$endgroup$
– Mathmo123
Jul 20 '14 at 18:23
$begingroup$
Isn't a prime number positive by definition? Thanks Shaun I tried to make it better
$endgroup$
– user121591
Jul 20 '14 at 18:25
1
$begingroup$
I don't think your first line is correct. Suppose they are all greater than two. Then indeed all of the products of the form $pq-1$ are even. But that isn't a problem (even numbers can have odd prime divisors), and it doesn't immediately imply anything interesting.
$endgroup$
– Potato
Jul 20 '14 at 18:35
$begingroup$
By the way, I think you can piece a solution together from the comments here: nrich.maths.org/discus/messages/153904/149403.html?1281107322
$endgroup$
– Potato
Jul 20 '14 at 18:40
|
show 4 more comments
$begingroup$
Here's a MathJax tutorial :)
$endgroup$
– Shaun
Jul 20 '14 at 18:21
$begingroup$
Are negative primes allowed?
$endgroup$
– Mathmo123
Jul 20 '14 at 18:23
$begingroup$
Isn't a prime number positive by definition? Thanks Shaun I tried to make it better
$endgroup$
– user121591
Jul 20 '14 at 18:25
1
$begingroup$
I don't think your first line is correct. Suppose they are all greater than two. Then indeed all of the products of the form $pq-1$ are even. But that isn't a problem (even numbers can have odd prime divisors), and it doesn't immediately imply anything interesting.
$endgroup$
– Potato
Jul 20 '14 at 18:35
$begingroup$
By the way, I think you can piece a solution together from the comments here: nrich.maths.org/discus/messages/153904/149403.html?1281107322
$endgroup$
– Potato
Jul 20 '14 at 18:40
$begingroup$
Here's a MathJax tutorial :)
$endgroup$
– Shaun
Jul 20 '14 at 18:21
$begingroup$
Here's a MathJax tutorial :)
$endgroup$
– Shaun
Jul 20 '14 at 18:21
$begingroup$
Are negative primes allowed?
$endgroup$
– Mathmo123
Jul 20 '14 at 18:23
$begingroup$
Are negative primes allowed?
$endgroup$
– Mathmo123
Jul 20 '14 at 18:23
$begingroup$
Isn't a prime number positive by definition? Thanks Shaun I tried to make it better
$endgroup$
– user121591
Jul 20 '14 at 18:25
$begingroup$
Isn't a prime number positive by definition? Thanks Shaun I tried to make it better
$endgroup$
– user121591
Jul 20 '14 at 18:25
1
1
$begingroup$
I don't think your first line is correct. Suppose they are all greater than two. Then indeed all of the products of the form $pq-1$ are even. But that isn't a problem (even numbers can have odd prime divisors), and it doesn't immediately imply anything interesting.
$endgroup$
– Potato
Jul 20 '14 at 18:35
$begingroup$
I don't think your first line is correct. Suppose they are all greater than two. Then indeed all of the products of the form $pq-1$ are even. But that isn't a problem (even numbers can have odd prime divisors), and it doesn't immediately imply anything interesting.
$endgroup$
– Potato
Jul 20 '14 at 18:35
$begingroup$
By the way, I think you can piece a solution together from the comments here: nrich.maths.org/discus/messages/153904/149403.html?1281107322
$endgroup$
– Potato
Jul 20 '14 at 18:40
$begingroup$
By the way, I think you can piece a solution together from the comments here: nrich.maths.org/discus/messages/153904/149403.html?1281107322
$endgroup$
– Potato
Jul 20 '14 at 18:40
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
The divisibility criteria ensure that $pqr$ divides
$$(pq-1)(pr-1)(qr-1)=p^2q^2r^2-p^2qr-pq^2r-pqr^2+pq+pr+qr-1,$$
and hence that $pqrmid pq+pr+qr-1$. Without loss of generality $pleq qleq r$ and so
$$3qr-1geq pq+pr+qr-1geq pqr,$$
which shows that $p<3$ and so $p=2$. Then $2qrmid2q+2r+qr-1$ and so $q$ and $r$ are odd, and moreover
$$2qrleq 2q+2r+qr-1,$$
or equivalently $(q-2)(r-2)leq3$. This shows that $q=3$ and $rin3,5$, and a quick check shows that $r=3$ does not work and that $r=5$ does, so $p,q,r=2,3,5$ and so $pqr=30$.
answered Mar 29 at 22:24
ServaesServaes
30k342101
$endgroup$
add a comment |
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$begingroup$
The divisibility criteria ensure that $pqr$ divides
$$(pq-1)(pr-1)(qr-1)=p^2q^2r^2-p^2qr-pq^2r-pqr^2+pq+pr+qr-1,$$
and hence that $pqrmid pq+pr+qr-1$. Without loss of generality $pleq qleq r$ and so
$$3qr-1geq pq+pr+qr-1geq pqr,$$
which shows that $p<3$ and so $p=2$. Then $2qrmid2q+2r+qr-1$ and so $q$ and $r$ are odd, and moreover
$$2qrleq 2q+2r+qr-1,$$
or equivalently $(q-2)(r-2)leq3$. This shows that $q=3$ and $rin3,5$, and a quick check shows that $r=3$ does not work and that $r=5$ does, so $p,q,r=2,3,5$ and so $pqr=30$.
answered Mar 29 at 22:24
ServaesServaes
30k342101
$endgroup$
add a comment |
$begingroup$
The divisibility criteria ensure that $pqr$ divides
$$(pq-1)(pr-1)(qr-1)=p^2q^2r^2-p^2qr-pq^2r-pqr^2+pq+pr+qr-1,$$
and hence that $pqrmid pq+pr+qr-1$. Without loss of generality $pleq qleq r$ and so
$$3qr-1geq pq+pr+qr-1geq pqr,$$
which shows that $p<3$ and so $p=2$. Then $2qrmid2q+2r+qr-1$ and so $q$ and $r$ are odd, and moreover
$$2qrleq 2q+2r+qr-1,$$
or equivalently $(q-2)(r-2)leq3$. This shows that $q=3$ and $rin3,5$, and a quick check shows that $r=3$ does not work and that $r=5$ does, so $p,q,r=2,3,5$ and so $pqr=30$.
answered Mar 29 at 22:24
ServaesServaes
30k342101
$endgroup$
add a comment |
$begingroup$
The divisibility criteria ensure that $pqr$ divides
$$(pq-1)(pr-1)(qr-1)=p^2q^2r^2-p^2qr-pq^2r-pqr^2+pq+pr+qr-1,$$
and hence that $pqrmid pq+pr+qr-1$. Without loss of generality $pleq qleq r$ and so
$$3qr-1geq pq+pr+qr-1geq pqr,$$
which shows that $p<3$ and so $p=2$. Then $2qrmid2q+2r+qr-1$ and so $q$ and $r$ are odd, and moreover
$$2qrleq 2q+2r+qr-1,$$
or equivalently $(q-2)(r-2)leq3$. This shows that $q=3$ and $rin3,5$, and a quick check shows that $r=3$ does not work and that $r=5$ does, so $p,q,r=2,3,5$ and so $pqr=30$.
answered Mar 29 at 22:24
ServaesServaes
30k342101
$endgroup$
The divisibility criteria ensure that $pqr$ divides
$$(pq-1)(pr-1)(qr-1)=p^2q^2r^2-p^2qr-pq^2r-pqr^2+pq+pr+qr-1,$$
and hence that $pqrmid pq+pr+qr-1$. Without loss of generality $pleq qleq r$ and so
$$3qr-1geq pq+pr+qr-1geq pqr,$$
which shows that $p<3$ and so $p=2$. Then $2qrmid2q+2r+qr-1$ and so $q$ and $r$ are odd, and moreover
$$2qrleq 2q+2r+qr-1,$$
or equivalently $(q-2)(r-2)leq3$. This shows that $q=3$ and $rin3,5$, and a quick check shows that $r=3$ does not work and that $r=5$ does, so $p,q,r=2,3,5$ and so $pqr=30$.
answered Mar 29 at 22:24
ServaesServaes
30k342101
answered Mar 29 at 22:24
ServaesServaes
30k342101
answered Mar 29 at 22:24
ServaesServaes
30k342101
answered Mar 29 at 22:24
ServaesServaes
30k342101
30k342101
add a comment |
add a comment |
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$begingroup$
Here's a MathJax tutorial :)
$endgroup$
– Shaun
Jul 20 '14 at 18:21
$begingroup$
Are negative primes allowed?
$endgroup$
– Mathmo123
Jul 20 '14 at 18:23
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Isn't a prime number positive by definition? Thanks Shaun I tried to make it better
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– user121591
Jul 20 '14 at 18:25
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I don't think your first line is correct. Suppose they are all greater than two. Then indeed all of the products of the form $pq-1$ are even. But that isn't a problem (even numbers can have odd prime divisors), and it doesn't immediately imply anything interesting.
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– Potato
Jul 20 '14 at 18:35
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By the way, I think you can piece a solution together from the comments here: nrich.maths.org/discus/messages/153904/149403.html?1281107322
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– Potato
Jul 20 '14 at 18:40