Mixing
Find all positive triples of positive integers a, b, c so that $frac {a+1}{b}$ , $frac {b+1}{c}$, $frac...
$begingroup$
Find all positive triples of positive integers a, b, c so that $frac {a+1}{b}$ , $frac {b+1}{c}$, $frac {c+1}{a}$ are also integers.
WLOG, let a$leqq bleqq c$,
elementary-number-theory
asked Jan 19 at 22:42
John AburiJohn Aburi
142
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add a comment |
$begingroup$
Find all positive triples of positive integers a, b, c so that $frac {a+1}{b}$ , $frac {b+1}{c}$, $frac {c+1}{a}$ are also integers.
WLOG, let a$leqq bleqq c$,
elementary-number-theory
asked Jan 19 at 22:42
John AburiJohn Aburi
142
$endgroup$
1
$begingroup$
Did you find any such triples?
$endgroup$
– coffeemath
Jan 19 at 22:44
1
$begingroup$
Are you sure that is WLOG? The condition is not invariant under arbitrary permutations of $a,b,c$ but only under 3-cycles. A priori, there might be a solution where, as you go around the cycle, there are two decreasing and one increasing step.
$endgroup$
– Henning Makholm
Jan 19 at 22:51
$begingroup$
It's wolog that $a = min(a,b,c)$ but it's not wolog that $mid(a,b,c)|min(a,b,c)+1$
$endgroup$
– fleablood
Jan 19 at 23:10
add a comment |
$begingroup$
Find all positive triples of positive integers a, b, c so that $frac {a+1}{b}$ , $frac {b+1}{c}$, $frac {c+1}{a}$ are also integers.
WLOG, let a$leqq bleqq c$,
elementary-number-theory
asked Jan 19 at 22:42
John AburiJohn Aburi
142
$endgroup$
Find all positive triples of positive integers a, b, c so that $frac {a+1}{b}$ , $frac {b+1}{c}$, $frac {c+1}{a}$ are also integers.
WLOG, let a$leqq bleqq c$,
elementary-number-theory
elementary-number-theory
asked Jan 19 at 22:42
John AburiJohn Aburi
142
asked Jan 19 at 22:42
John AburiJohn Aburi
142
asked Jan 19 at 22:42
John AburiJohn Aburi
142
asked Jan 19 at 22:42
John AburiJohn Aburi
142
asked Jan 19 at 22:42
John AburiJohn Aburi
142
142
1
$begingroup$
Did you find any such triples?
$endgroup$
– coffeemath
Jan 19 at 22:44
1
$begingroup$
Are you sure that is WLOG? The condition is not invariant under arbitrary permutations of $a,b,c$ but only under 3-cycles. A priori, there might be a solution where, as you go around the cycle, there are two decreasing and one increasing step.
$endgroup$
– Henning Makholm
Jan 19 at 22:51
$begingroup$
It's wolog that $a = min(a,b,c)$ but it's not wolog that $mid(a,b,c)|min(a,b,c)+1$
$endgroup$
– fleablood
Jan 19 at 23:10
add a comment |
1
$begingroup$
Did you find any such triples?
$endgroup$
– coffeemath
Jan 19 at 22:44
1
$begingroup$
Are you sure that is WLOG? The condition is not invariant under arbitrary permutations of $a,b,c$ but only under 3-cycles. A priori, there might be a solution where, as you go around the cycle, there are two decreasing and one increasing step.
$endgroup$
– Henning Makholm
Jan 19 at 22:51
$begingroup$
It's wolog that $a = min(a,b,c)$ but it's not wolog that $mid(a,b,c)|min(a,b,c)+1$
$endgroup$
– fleablood
Jan 19 at 23:10
1
1
$begingroup$
Did you find any such triples?
$endgroup$
– coffeemath
Jan 19 at 22:44
$begingroup$
Did you find any such triples?
$endgroup$
– coffeemath
Jan 19 at 22:44
1
1
$begingroup$
Are you sure that is WLOG? The condition is not invariant under arbitrary permutations of $a,b,c$ but only under 3-cycles. A priori, there might be a solution where, as you go around the cycle, there are two decreasing and one increasing step.
$endgroup$
– Henning Makholm
Jan 19 at 22:51
$begingroup$
Are you sure that is WLOG? The condition is not invariant under arbitrary permutations of $a,b,c$ but only under 3-cycles. A priori, there might be a solution where, as you go around the cycle, there are two decreasing and one increasing step.
$endgroup$
– Henning Makholm
Jan 19 at 22:51
$begingroup$
It's wolog that $a = min(a,b,c)$ but it's not wolog that $mid(a,b,c)|min(a,b,c)+1$
$endgroup$
– fleablood
Jan 19 at 23:10
$begingroup$
It's wolog that $a = min(a,b,c)$ but it's not wolog that $mid(a,b,c)|min(a,b,c)+1$
$endgroup$
– fleablood
Jan 19 at 23:10
add a comment |
3 Answers
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Hint: given $a le b$ and $frac {a+1}b$ is an integer, you must have $b=a+1$ or $a=b=1$.
answered Jan 19 at 22:45
Ross MillikanRoss Millikan
298k23198371
$endgroup$
3
$begingroup$
Or $a = b = 1$.
$endgroup$
– Henning Makholm
Jan 19 at 22:49
$begingroup$
@HenningMakholm: Good point. Thanks.
$endgroup$
– Ross Millikan
Jan 19 at 22:51
add a comment |
$begingroup$
If any two of $a,b,c$ are equal, then wlog. $a=b$. As $frac{b+1}{a}=1+frac1a$ is an integer, we conclude $a=b=1$. The remaining conditions are that $frac{c+1}{1}$ and $frac 2c$ are integers, which lead us to the solutions
$$(1,1,1),qquad (1,1,2) $$
(and cyclic permutations of the latter).
So assume $a,b,c $ are pairwise different. By cyclic permutation, we may assume wlog that $a<b<c$ or that $a>b>c$.
In the first case, $0<frac{a+1}{b}le frac bb=1$ and hence $a+1=b$. Likewise, $b+1=c$. Then the last integer is $frac{c+1}a=frac{a+3}a=1+frac 3a$ and we must have $a=1$ or $a=3$, whic gives us the solutions
$$(1,2,3),qquad (3,4,5) $$
(and cyclic permutations).
In the case $a>b>c$, we instead have that $0<frac{c+1}{a}le frac{c+1}{c+2}<1$, not an integer. So this case does not produce additional solutions.
answered Jan 19 at 23:05
Hagen von EitzenHagen von Eitzen
281k23272505
$endgroup$
add a comment |
$begingroup$
If $a le b$ but $b|a+1$ then $b le a+1$ so either $a = b$ or $b=a+1$.
If $a = b$ then $b|b+1$ and $a=b=1$. The you $c|b+1=2$ and $b|c+1$. So either $c=1$ or $c = 2$.
So so far we have $(1,1,1)$ or $(1,1,2)$
If $b = a+1$ then $c|a+2$ so $c le a+2$ but $a < b=a+1 le cle a+2 $ so either $c = b = a+1$ or $c= a+2$.
If $c= b =a+1$ then we have $b|a+1 = b$ and $c=b|b+1$ and $a=b-1|c+1=b+1$. So $b=1$ but then $a=0$ and that's a contradiction.
If $c=a+2$ and $b= a+1$ we have: $b|a+1=b$; $c|b+1 =c$ and $a|c+1 = a+3$. This means $a|3$ and $a =1$ or $a =3$.
So we have $(1,2,3)$ or $(3,4,5)$.
To double check:
$(a,b,c) = (1,1,1)implies $ $b|a+1, c|b+1, a|c+1$ all translate to $1|2$. which is true
$(a,b,c) = (1,1,2) implies b=1|a+1=2; c=2|b+1=2; a=1|c+1=3$. All true.
$(a,b,c) = (1,2,3)implies b=2|a+1=2; c=3|b+1=3; a=1|c+1 = 4$. All true.
$(a,b,c) = (3,4,5)implies b=4|a+1=4; c=5|b+1=5; a=3|c+1=6$. All true.
answered Jan 19 at 23:06
fleabloodfleablood
71.7k22686
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add a comment |
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3 Answers
3
active
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votes
3 Answers
3
active
oldest
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active
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votes
$begingroup$
Hint: given $a le b$ and $frac {a+1}b$ is an integer, you must have $b=a+1$ or $a=b=1$.
answered Jan 19 at 22:45
Ross MillikanRoss Millikan
298k23198371
$endgroup$
3
$begingroup$
Or $a = b = 1$.
$endgroup$
– Henning Makholm
Jan 19 at 22:49
$begingroup$
@HenningMakholm: Good point. Thanks.
$endgroup$
– Ross Millikan
Jan 19 at 22:51
add a comment |
$begingroup$
Hint: given $a le b$ and $frac {a+1}b$ is an integer, you must have $b=a+1$ or $a=b=1$.
answered Jan 19 at 22:45
Ross MillikanRoss Millikan
298k23198371
$endgroup$
3
$begingroup$
Or $a = b = 1$.
$endgroup$
– Henning Makholm
Jan 19 at 22:49
$begingroup$
@HenningMakholm: Good point. Thanks.
$endgroup$
– Ross Millikan
Jan 19 at 22:51
add a comment |
$begingroup$
Hint: given $a le b$ and $frac {a+1}b$ is an integer, you must have $b=a+1$ or $a=b=1$.
answered Jan 19 at 22:45
Ross MillikanRoss Millikan
298k23198371
$endgroup$
Hint: given $a le b$ and $frac {a+1}b$ is an integer, you must have $b=a+1$ or $a=b=1$.
answered Jan 19 at 22:45
Ross MillikanRoss Millikan
298k23198371
edited Jan 19 at 22:51
answered Jan 19 at 22:45
Ross MillikanRoss Millikan
298k23198371
answered Jan 19 at 22:45
Ross MillikanRoss Millikan
298k23198371
answered Jan 19 at 22:45
Ross MillikanRoss Millikan
298k23198371
298k23198371
3
$begingroup$
Or $a = b = 1$.
$endgroup$
– Henning Makholm
Jan 19 at 22:49
$begingroup$
@HenningMakholm: Good point. Thanks.
$endgroup$
– Ross Millikan
Jan 19 at 22:51
add a comment |
3
$begingroup$
Or $a = b = 1$.
$endgroup$
– Henning Makholm
Jan 19 at 22:49
$begingroup$
@HenningMakholm: Good point. Thanks.
$endgroup$
– Ross Millikan
Jan 19 at 22:51
3
3
$begingroup$
Or $a = b = 1$.
$endgroup$
– Henning Makholm
Jan 19 at 22:49
$begingroup$
Or $a = b = 1$.
$endgroup$
– Henning Makholm
Jan 19 at 22:49
$begingroup$
@HenningMakholm: Good point. Thanks.
$endgroup$
– Ross Millikan
Jan 19 at 22:51
$begingroup$
@HenningMakholm: Good point. Thanks.
$endgroup$
– Ross Millikan
Jan 19 at 22:51
add a comment |
$begingroup$
If any two of $a,b,c$ are equal, then wlog. $a=b$. As $frac{b+1}{a}=1+frac1a$ is an integer, we conclude $a=b=1$. The remaining conditions are that $frac{c+1}{1}$ and $frac 2c$ are integers, which lead us to the solutions
$$(1,1,1),qquad (1,1,2) $$
(and cyclic permutations of the latter).
So assume $a,b,c $ are pairwise different. By cyclic permutation, we may assume wlog that $a<b<c$ or that $a>b>c$.
In the first case, $0<frac{a+1}{b}le frac bb=1$ and hence $a+1=b$. Likewise, $b+1=c$. Then the last integer is $frac{c+1}a=frac{a+3}a=1+frac 3a$ and we must have $a=1$ or $a=3$, whic gives us the solutions
$$(1,2,3),qquad (3,4,5) $$
(and cyclic permutations).
In the case $a>b>c$, we instead have that $0<frac{c+1}{a}le frac{c+1}{c+2}<1$, not an integer. So this case does not produce additional solutions.
answered Jan 19 at 23:05
Hagen von EitzenHagen von Eitzen
281k23272505
$endgroup$
add a comment |
$begingroup$
If any two of $a,b,c$ are equal, then wlog. $a=b$. As $frac{b+1}{a}=1+frac1a$ is an integer, we conclude $a=b=1$. The remaining conditions are that $frac{c+1}{1}$ and $frac 2c$ are integers, which lead us to the solutions
$$(1,1,1),qquad (1,1,2) $$
(and cyclic permutations of the latter).
So assume $a,b,c $ are pairwise different. By cyclic permutation, we may assume wlog that $a<b<c$ or that $a>b>c$.
In the first case, $0<frac{a+1}{b}le frac bb=1$ and hence $a+1=b$. Likewise, $b+1=c$. Then the last integer is $frac{c+1}a=frac{a+3}a=1+frac 3a$ and we must have $a=1$ or $a=3$, whic gives us the solutions
$$(1,2,3),qquad (3,4,5) $$
(and cyclic permutations).
In the case $a>b>c$, we instead have that $0<frac{c+1}{a}le frac{c+1}{c+2}<1$, not an integer. So this case does not produce additional solutions.
answered Jan 19 at 23:05
Hagen von EitzenHagen von Eitzen
281k23272505
$endgroup$
add a comment |
$begingroup$
If any two of $a,b,c$ are equal, then wlog. $a=b$. As $frac{b+1}{a}=1+frac1a$ is an integer, we conclude $a=b=1$. The remaining conditions are that $frac{c+1}{1}$ and $frac 2c$ are integers, which lead us to the solutions
$$(1,1,1),qquad (1,1,2) $$
(and cyclic permutations of the latter).
So assume $a,b,c $ are pairwise different. By cyclic permutation, we may assume wlog that $a<b<c$ or that $a>b>c$.
In the first case, $0<frac{a+1}{b}le frac bb=1$ and hence $a+1=b$. Likewise, $b+1=c$. Then the last integer is $frac{c+1}a=frac{a+3}a=1+frac 3a$ and we must have $a=1$ or $a=3$, whic gives us the solutions
$$(1,2,3),qquad (3,4,5) $$
(and cyclic permutations).
In the case $a>b>c$, we instead have that $0<frac{c+1}{a}le frac{c+1}{c+2}<1$, not an integer. So this case does not produce additional solutions.
answered Jan 19 at 23:05
Hagen von EitzenHagen von Eitzen
281k23272505
$endgroup$
If any two of $a,b,c$ are equal, then wlog. $a=b$. As $frac{b+1}{a}=1+frac1a$ is an integer, we conclude $a=b=1$. The remaining conditions are that $frac{c+1}{1}$ and $frac 2c$ are integers, which lead us to the solutions
$$(1,1,1),qquad (1,1,2) $$
(and cyclic permutations of the latter).
So assume $a,b,c $ are pairwise different. By cyclic permutation, we may assume wlog that $a<b<c$ or that $a>b>c$.
In the first case, $0<frac{a+1}{b}le frac bb=1$ and hence $a+1=b$. Likewise, $b+1=c$. Then the last integer is $frac{c+1}a=frac{a+3}a=1+frac 3a$ and we must have $a=1$ or $a=3$, whic gives us the solutions
$$(1,2,3),qquad (3,4,5) $$
(and cyclic permutations).
In the case $a>b>c$, we instead have that $0<frac{c+1}{a}le frac{c+1}{c+2}<1$, not an integer. So this case does not produce additional solutions.
answered Jan 19 at 23:05
Hagen von EitzenHagen von Eitzen
281k23272505
answered Jan 19 at 23:05
Hagen von EitzenHagen von Eitzen
281k23272505
answered Jan 19 at 23:05
Hagen von EitzenHagen von Eitzen
281k23272505
answered Jan 19 at 23:05
Hagen von EitzenHagen von Eitzen
281k23272505
281k23272505
add a comment |
add a comment |
$begingroup$
If $a le b$ but $b|a+1$ then $b le a+1$ so either $a = b$ or $b=a+1$.
If $a = b$ then $b|b+1$ and $a=b=1$. The you $c|b+1=2$ and $b|c+1$. So either $c=1$ or $c = 2$.
So so far we have $(1,1,1)$ or $(1,1,2)$
If $b = a+1$ then $c|a+2$ so $c le a+2$ but $a < b=a+1 le cle a+2 $ so either $c = b = a+1$ or $c= a+2$.
If $c= b =a+1$ then we have $b|a+1 = b$ and $c=b|b+1$ and $a=b-1|c+1=b+1$. So $b=1$ but then $a=0$ and that's a contradiction.
If $c=a+2$ and $b= a+1$ we have: $b|a+1=b$; $c|b+1 =c$ and $a|c+1 = a+3$. This means $a|3$ and $a =1$ or $a =3$.
So we have $(1,2,3)$ or $(3,4,5)$.
To double check:
$(a,b,c) = (1,1,1)implies $ $b|a+1, c|b+1, a|c+1$ all translate to $1|2$. which is true
$(a,b,c) = (1,1,2) implies b=1|a+1=2; c=2|b+1=2; a=1|c+1=3$. All true.
$(a,b,c) = (1,2,3)implies b=2|a+1=2; c=3|b+1=3; a=1|c+1 = 4$. All true.
$(a,b,c) = (3,4,5)implies b=4|a+1=4; c=5|b+1=5; a=3|c+1=6$. All true.
answered Jan 19 at 23:06
fleabloodfleablood
71.7k22686
$endgroup$
add a comment |
$begingroup$
If $a le b$ but $b|a+1$ then $b le a+1$ so either $a = b$ or $b=a+1$.
If $a = b$ then $b|b+1$ and $a=b=1$. The you $c|b+1=2$ and $b|c+1$. So either $c=1$ or $c = 2$.
So so far we have $(1,1,1)$ or $(1,1,2)$
If $b = a+1$ then $c|a+2$ so $c le a+2$ but $a < b=a+1 le cle a+2 $ so either $c = b = a+1$ or $c= a+2$.
If $c= b =a+1$ then we have $b|a+1 = b$ and $c=b|b+1$ and $a=b-1|c+1=b+1$. So $b=1$ but then $a=0$ and that's a contradiction.
If $c=a+2$ and $b= a+1$ we have: $b|a+1=b$; $c|b+1 =c$ and $a|c+1 = a+3$. This means $a|3$ and $a =1$ or $a =3$.
So we have $(1,2,3)$ or $(3,4,5)$.
To double check:
$(a,b,c) = (1,1,1)implies $ $b|a+1, c|b+1, a|c+1$ all translate to $1|2$. which is true
$(a,b,c) = (1,1,2) implies b=1|a+1=2; c=2|b+1=2; a=1|c+1=3$. All true.
$(a,b,c) = (1,2,3)implies b=2|a+1=2; c=3|b+1=3; a=1|c+1 = 4$. All true.
$(a,b,c) = (3,4,5)implies b=4|a+1=4; c=5|b+1=5; a=3|c+1=6$. All true.
answered Jan 19 at 23:06
fleabloodfleablood
71.7k22686
$endgroup$
add a comment |
$begingroup$
If $a le b$ but $b|a+1$ then $b le a+1$ so either $a = b$ or $b=a+1$.
If $a = b$ then $b|b+1$ and $a=b=1$. The you $c|b+1=2$ and $b|c+1$. So either $c=1$ or $c = 2$.
So so far we have $(1,1,1)$ or $(1,1,2)$
If $b = a+1$ then $c|a+2$ so $c le a+2$ but $a < b=a+1 le cle a+2 $ so either $c = b = a+1$ or $c= a+2$.
If $c= b =a+1$ then we have $b|a+1 = b$ and $c=b|b+1$ and $a=b-1|c+1=b+1$. So $b=1$ but then $a=0$ and that's a contradiction.
If $c=a+2$ and $b= a+1$ we have: $b|a+1=b$; $c|b+1 =c$ and $a|c+1 = a+3$. This means $a|3$ and $a =1$ or $a =3$.
So we have $(1,2,3)$ or $(3,4,5)$.
To double check:
$(a,b,c) = (1,1,1)implies $ $b|a+1, c|b+1, a|c+1$ all translate to $1|2$. which is true
$(a,b,c) = (1,1,2) implies b=1|a+1=2; c=2|b+1=2; a=1|c+1=3$. All true.
$(a,b,c) = (1,2,3)implies b=2|a+1=2; c=3|b+1=3; a=1|c+1 = 4$. All true.
$(a,b,c) = (3,4,5)implies b=4|a+1=4; c=5|b+1=5; a=3|c+1=6$. All true.
answered Jan 19 at 23:06
fleabloodfleablood
71.7k22686
$endgroup$
If $a le b$ but $b|a+1$ then $b le a+1$ so either $a = b$ or $b=a+1$.
If $a = b$ then $b|b+1$ and $a=b=1$. The you $c|b+1=2$ and $b|c+1$. So either $c=1$ or $c = 2$.
So so far we have $(1,1,1)$ or $(1,1,2)$
If $b = a+1$ then $c|a+2$ so $c le a+2$ but $a < b=a+1 le cle a+2 $ so either $c = b = a+1$ or $c= a+2$.
If $c= b =a+1$ then we have $b|a+1 = b$ and $c=b|b+1$ and $a=b-1|c+1=b+1$. So $b=1$ but then $a=0$ and that's a contradiction.
If $c=a+2$ and $b= a+1$ we have: $b|a+1=b$; $c|b+1 =c$ and $a|c+1 = a+3$. This means $a|3$ and $a =1$ or $a =3$.
So we have $(1,2,3)$ or $(3,4,5)$.
To double check:
$(a,b,c) = (1,1,1)implies $ $b|a+1, c|b+1, a|c+1$ all translate to $1|2$. which is true
$(a,b,c) = (1,1,2) implies b=1|a+1=2; c=2|b+1=2; a=1|c+1=3$. All true.
$(a,b,c) = (1,2,3)implies b=2|a+1=2; c=3|b+1=3; a=1|c+1 = 4$. All true.
$(a,b,c) = (3,4,5)implies b=4|a+1=4; c=5|b+1=5; a=3|c+1=6$. All true.
answered Jan 19 at 23:06
fleabloodfleablood
71.7k22686
answered Jan 19 at 23:06
fleabloodfleablood
71.7k22686
answered Jan 19 at 23:06
fleabloodfleablood
71.7k22686
answered Jan 19 at 23:06
fleabloodfleablood
71.7k22686
71.7k22686
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1
$begingroup$
Did you find any such triples?
$endgroup$
– coffeemath
Jan 19 at 22:44
1
$begingroup$
Are you sure that is WLOG? The condition is not invariant under arbitrary permutations of $a,b,c$ but only under 3-cycles. A priori, there might be a solution where, as you go around the cycle, there are two decreasing and one increasing step.
$endgroup$
– Henning Makholm
Jan 19 at 22:51
$begingroup$
It's wolog that $a = min(a,b,c)$ but it's not wolog that $mid(a,b,c)|min(a,b,c)+1$
$endgroup$
– fleablood
Jan 19 at 23:10