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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 $b$.

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    $\begingroup$ 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? $\endgroup$ Commented Mar 24, 2019 at 12:59
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    $\begingroup$ Unless $b,c$ are prime it is not true that $b,c$ must divide one of $a+1,a-1$. $\endgroup$
    – lulu
    Commented Mar 24, 2019 at 13:03
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    $\begingroup$ 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
    Commented Mar 24, 2019 at 13:06
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    $\begingroup$ 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
    Commented Mar 24, 2019 at 13:29
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    $\begingroup$ Those triples include $(n^2-1,n^3-2n,n^4-3n^2+1)$ $\endgroup$
    – Empy2
    Commented Mar 24, 2019 at 13:59

1 Answer 1

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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.

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