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

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.

• Source of this question, please? Apr 4, 2020 at 4:21
• It's my homework...? Apr 4, 2020 at 5:19
• Are you permitted to post your homework here? Apr 4, 2020 at 8:54
• I mean, does your teacher permit you to post your homework here? Apr 4, 2020 at 11:52
• In short, you are posting your homework here, without permission from the person assigning the homework. Apr 4, 2020 at 22:07

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.

• this makes me depresive +1 Apr 4, 2020 at 16:30

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.