Find all positive integers $n$ such that $36^n - 6$ is the product of three consecutive naturals.

Question

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.

Answer 1

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.

Answer 2

Say $d= \gcd (m - 6, m^2 + 6m + 35)$ then $$m\equiv _d 6$$ and $$m^2+6m+35\equiv _d 0\implies 107 \equiv_d 0$$

so $d\in \{1,107\}$. If $d=107$ then $107\mid 6^n$ which is impossible, so $d=1$.

• Case $1$: $$ m-6 = 2^{2n} \;\;\;\wedge \;\;\;m^2+6m+35 = 3^{2n}$$ Second equation can be rewriten as $$(m+3)^2-3^{2n}= 26$$ which is easy to solve...

• Case $2$ $$ m-6 = 3^{2n} \;\;\;\wedge \;\;\;m^2+6m+35 = 2^{2n}$$ Second equation can be rewriten as $$(m+3)^2-2^{2n}= 26$$ which is again easy to solve...

• Case $3$ $$ m-6 = 1 \;\;\;\wedge \;\;\;m^2+6m+35 = 6^{2n}$$ So $m=7$ and ...

• Case $4$ $$ m-6 =6^{2n} \;\;\;\wedge \;\;\;m^2+6m+35 = 1$$ and no solution to second equation.