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Existence of 3 natural numbers that divide each other when squared and have 1 taken away from them

2

1

$begingroup$

Does there exist natural numbers, $a,b,c > 1$, such that;

$a^2 - 1$ is divisible by $b$ and $c$,
$b^2 - 1$ is divisible by $a$ and $c$ and
$c^2 - 1$ is divisible by $a$ and $b$.

number-theory natural-numbers

share|cite|improve this question

asked Mar 24 at 12:46

jjhhbb9jjhhbb9

183

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hello and welcome to math.stackexchange. This is a nice question. What are your thoughts? What have you tried? Any results from searching for examples?
  $endgroup$
  – Hans Engler
  Mar 24 at 12:59
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Unless $b,c$ are prime it is not true that $b,c$ must divide one of $a+1,a-1$.
  $endgroup$
  – lulu
  Mar 24 at 13:03
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  If you think that the first part of the problem statement implies that $b$ and $c$ must each divide $a-1$ or $a+1$, and similarly for the other two parts, I think you will quickly find that it can't work that way.
  $endgroup$
  – David K
  Mar 24 at 13:06
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The only triples with five out of six; and all three below 1000, are: $(3,4,5),(3,7,8),(8,21,55),(24,115,551),(15,56,209)$
  $endgroup$
  – Empy2
  Mar 24 at 13:29
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Those triples include $(n^2-1,n^3-2n,n^4-3n^2+1)$
  $endgroup$
  – Empy2
  Mar 24 at 13:59
  

  
  
  
  

 | 
show 3 more comments

2

1

$begingroup$

Does there exist natural numbers, $a,b,c > 1$, such that;

$a^2 - 1$ is divisible by $b$ and $c$,
$b^2 - 1$ is divisible by $a$ and $c$ and
$c^2 - 1$ is divisible by $a$ and $b$.

number-theory natural-numbers

share|cite|improve this question

asked Mar 24 at 12:46

jjhhbb9jjhhbb9

183

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hello and welcome to math.stackexchange. This is a nice question. What are your thoughts? What have you tried? Any results from searching for examples?
  $endgroup$
  – Hans Engler
  Mar 24 at 12:59
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Unless $b,c$ are prime it is not true that $b,c$ must divide one of $a+1,a-1$.
  $endgroup$
  – lulu
  Mar 24 at 13:03
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  If you think that the first part of the problem statement implies that $b$ and $c$ must each divide $a-1$ or $a+1$, and similarly for the other two parts, I think you will quickly find that it can't work that way.
  $endgroup$
  – David K
  Mar 24 at 13:06
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The only triples with five out of six; and all three below 1000, are: $(3,4,5),(3,7,8),(8,21,55),(24,115,551),(15,56,209)$
  $endgroup$
  – Empy2
  Mar 24 at 13:29
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Those triples include $(n^2-1,n^3-2n,n^4-3n^2+1)$
  $endgroup$
  – Empy2
  Mar 24 at 13:59
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

Does there exist natural numbers, $a,b,c > 1$, such that;

$a^2 - 1$ is divisible by $b$ and $c$,
$b^2 - 1$ is divisible by $a$ and $c$ and
$c^2 - 1$ is divisible by $a$ and $b$.

number-theory natural-numbers

share|cite|improve this question

asked Mar 24 at 12:46

jjhhbb9jjhhbb9

183

$endgroup$

share|cite|improve this question

asked Mar 24 at 12:46

jjhhbb9jjhhbb9

183

share|cite|improve this question

asked Mar 24 at 12:46

jjhhbb9jjhhbb9

183

asked Mar 24 at 12:46

jjhhbb9jjhhbb9

183

asked Mar 24 at 12:46

jjhhbb9jjhhbb9

183

1 Answer
1

active

oldest

votes

7

$begingroup$

No such numbers exist.

First, observe that $a,b,c$ are pairwise coprime: For example, since $a$ divides $b^2-1$, any prime $p$ that divides $a$ also divides $b^2 - 1$; hence $p$ does not divide $b^2$ and also does not divide $b$. Thus $a$ and $b$ are coprime; and likewise for the other pairs.

Since $b,c$ are coprime and each divide $a^2 - 1$, so does their product $bc$. Since all the quantities are positive, that implies $bc le a^2 - 1 lt a^2$. So we have three strict inequalities:
$$
begin{align}
bc &lt a^2\
ac &lt b^2\
ab &lt c^2
end{align}
$$
Multiplying these inequalities together yields
$$
a^2 b^2 c^2 lt a^2 b^2 c^2
$$
which is impossible.

share|cite|improve this answer

answered Mar 24 at 15:08

FredHFredH

3,6851024

$endgroup$

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7

$begingroup$

No such numbers exist.

First, observe that $a,b,c$ are pairwise coprime: For example, since $a$ divides $b^2-1$, any prime $p$ that divides $a$ also divides $b^2 - 1$; hence $p$ does not divide $b^2$ and also does not divide $b$. Thus $a$ and $b$ are coprime; and likewise for the other pairs.

Since $b,c$ are coprime and each divide $a^2 - 1$, so does their product $bc$. Since all the quantities are positive, that implies $bc le a^2 - 1 lt a^2$. So we have three strict inequalities:
$$
begin{align}
bc &lt a^2\
ac &lt b^2\
ab &lt c^2
end{align}
$$
Multiplying these inequalities together yields
$$
a^2 b^2 c^2 lt a^2 b^2 c^2
$$
which is impossible.

share|cite|improve this answer

answered Mar 24 at 15:08

FredHFredH

3,6851024

$endgroup$

add a comment | 

7

$begingroup$

No such numbers exist.

First, observe that $a,b,c$ are pairwise coprime: For example, since $a$ divides $b^2-1$, any prime $p$ that divides $a$ also divides $b^2 - 1$; hence $p$ does not divide $b^2$ and also does not divide $b$. Thus $a$ and $b$ are coprime; and likewise for the other pairs.

Since $b,c$ are coprime and each divide $a^2 - 1$, so does their product $bc$. Since all the quantities are positive, that implies $bc le a^2 - 1 lt a^2$. So we have three strict inequalities:
$$
begin{align}
bc &lt a^2\
ac &lt b^2\
ab &lt c^2
end{align}
$$
Multiplying these inequalities together yields
$$
a^2 b^2 c^2 lt a^2 b^2 c^2
$$
which is impossible.

share|cite|improve this answer

answered Mar 24 at 15:08

FredHFredH

3,6851024

$endgroup$

add a comment | 

$begingroup$

No such numbers exist.

First, observe that $a,b,c$ are pairwise coprime: For example, since $a$ divides $b^2-1$, any prime $p$ that divides $a$ also divides $b^2 - 1$; hence $p$ does not divide $b^2$ and also does not divide $b$. Thus $a$ and $b$ are coprime; and likewise for the other pairs.

Since $b,c$ are coprime and each divide $a^2 - 1$, so does their product $bc$. Since all the quantities are positive, that implies $bc le a^2 - 1 lt a^2$. So we have three strict inequalities:
$$
begin{align}
bc &lt a^2\
ac &lt b^2\
ab &lt c^2
end{align}
$$
Multiplying these inequalities together yields
$$
a^2 b^2 c^2 lt a^2 b^2 c^2
$$
which is impossible.

share|cite|improve this answer

answered Mar 24 at 15:08

FredHFredH

3,6851024

$endgroup$

share|cite|improve this answer

answered Mar 24 at 15:08

FredHFredH

3,6851024

answered Mar 24 at 15:08

FredHFredH

3,6851024

answered Mar 24 at 15:08

FredHFredH

3,6851024