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

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.
