VMO

For any integer $k$, let $N_k$ be the number of ordered 6-tuple $(a,b,c,a^{\prime },b^{\prime },c^{\prime })$ that satisfy $$ab+a^{\prime }{b}^{\prime }\equiv bc+b^{\prime }{c}^{\prime }\equiv ca+c^{\prime }{a}^{\prime }\equiv 1(\mathrm{mod}k)$$ and $a,b,c,a^{\prime },b^{\prime },c^{\prime }\in \{0,1,\dots ,k-1\}$. By the Chinese Remainder Theorem, $$N_{mn}=N_n\times N_m\text{ if }\gcd (n,m)=1.$$ Therefore, in order to compute $N_{15}$, we only need to compute $N_3$ and $N_5$. We will compute $N_p$ for any $p$ prime. Fix a solution $(a,b,a^{\prime },b^{\prime })$ of the equation $ab+a^{\prime }{b}^{\prime }\equiv 1(\mathrm{mod}p)$, we compute the number of solutions of the following system $$bc+b^{\prime }{c}^{\prime }\equiv ca+c^{\prime }{a}^{\prime }\equiv 1(\mathrm{mod}p)(1)$$ We consider three cases

Suppose that $(a,a^{\prime })\ne t(b,b^{\prime })(\mathrm{mod}p)$ for any $t\in \{0,1,\dots ,p-1\}$. Then the system (1) has a unique solution $$c\equiv \frac{a^{\prime }-b^{\prime }}{a^{\prime }b-b^{\prime }a},c^{\prime }\equiv \frac{a-b}{a{b}^{\prime }-b{a}^{\prime }}(\mathrm{mod}p)$$ Suppose that $(a,a^{\prime })\equiv t(b,b^{\prime })(\mathrm{mod}p)$ for some $t\ne 1$. Then the system (1) has no solution. * Suppose that $(a,a^{\prime })\equiv (b,b^{\prime })(\mathrm{mod}p)$. Then the system (1) becomes a single equation $bc+b^{\prime }{c}^{\prime }\equiv 1(\mathrm{mod}p)$. Since $ba+b^{\prime }{a}^{\prime }\equiv 1(\mathrm{mod}p)$, we can assume that $b\ne 0$. Hence, for any choice of $c^{\prime }$, we have only one choice of $c\equiv (1-b^{\prime }{c}^{\prime })/b(\mathrm{mod}p)$. This implies that the system (1) has exactly $p$ solutions.

Let $T_p$ be the number of ordered tuples $(a,b,a^{\prime },b^{\prime })$ that satisfy $ab+a^{\prime }{b}^{\prime }\equiv 1(\mathrm{mod}p)$ and $a,b,a^{\prime },b^{\prime }\in \{0,1,\dots ,p-1\}$. For any pair $(a,a^{\prime })\ne (0,0)$, there are exactly $p$ pairs $(b,b^{\prime })$ satisfy the equation. Hence, $T_p=p(p^2-1)$.

Let $C_p(t)$ be the number of ordered pairs $(a,b)$ that satisfy $a^2+b^2\equiv t(\mathrm{mod}p)$ and $a,b\in \{0,1,\dots ,p-1\}$. From the above arguments, we have $$N_p=T_p-\sum _{t=1}^{p-1}{C}_p(t)+p{C}_p(1)=p(p^2-1)-p^2+C_p(0)+p{C}_p(1).$$ It is easy to get $C_3(0)=1,C_3(1)=4,C_5(0)=9,C_5(1)=4$, which implies that $N_3=28,N_5=124$ and $N_{15}=28\times 124=3472$.

Therefore, the number of ordered 6-tuple satisfying the given conditions is 3472. ☐