Find all positive integers $n$ such that $36^n – 6$ is the product of three consecutive naturals.
Let the second of the three naturals be $m$, $(m \in \mathbb Z^+, m \ge 1)$, we have that $(m – 1)m(m + 1) = m^3 – m$.
There doesn't exist natural $m$ such that $m^3 – m = 36^0 – 6 = -5$ or $m^3 – m = 36^1 – 6 = 30$.
Furthermore, $36^{3/2} – 6 = 6^3 – 6$, therefore $m = 6$ where $n = \dfrac{3}{2} \notin \mathbb N$ and $$m^3 – m – 210 = (m – 6)(m^2 + 6m + 35)$$
I believe that there don't exist any integer solutions $(m, n)$ such that $36^n – 6 = m^3 – m$, but I don't know how to prove so.
Best Answer
There are no solutions. This can be seen by examining the terms $\bmod 7$.
$36^n-6\equiv 1^n-(-1)=2 \bmod 7$
The product of three consecutive integers $\bmod 7$ will be $\equiv \{(0\cdot 1\cdot 2),(1\cdot 2\cdot 3),(2\cdot 3\cdot 4),(3\cdot 4\cdot 5),(4\cdot 5\cdot 6),(5\cdot 6\cdot 0),(6\cdot 0\cdot 1)\}=\{0,6,3,4,1,0,0\}$
Since the product of three consecutive integers is never $2\bmod 7$, the equation has no solutions.