# 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 $$b$$.

• 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? – Hans Engler Mar 24 at 12:59
• Unless $b,c$ are prime it is not true that $b,c$ must divide one of $a+1,a-1$. – lulu Mar 24 at 13:03
• 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. – David K Mar 24 at 13:06
• The only triples with five out of six; and all three below 1000, are: $(3,4,5),(3,7,8),(8,21,55),(24,115,551),(15,56,209)$ – Empy2 Mar 24 at 13:29
• Those triples include $(n^2-1,n^3-2n,n^4-3n^2+1)$ – Empy2 Mar 24 at 13:59

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.