The question is if there exist natural numbers $m$ and $n$ satisfying $(1) \;\; m(m+1)(m+2) = 36^n - 6$. The fact that $2 \cdot 3 \cdot 4 = 24 < 36^1 - 6 = 30 < 3 \cdot 4 \cdot 5 = 60$ means (1) has no solution when $n=1$. Hence $n \geq 2$. Moreover equation (1) is equivalent to $(2) \;\; (m+3)(m^2+2) = 36^n$. Let $d=GCD(m+3,m^2+2)$. Then $d \mid (-3)^2 + 2 = 9 + 2 = 11$, yielding $d \in \{1,11\}$. If $d=11$, then $11 \mid 36^n$ by (2), i.e. $11 \mid 36$. This contradiction means $d=1$, which according to (2) implies $(3) \;\; m+3 = a^n$, $(4) \;\, m^2+2 = b^n$, where $a$ and $b$ are coprime natural numbers s.t. $ab=36$. Assume $m$ is even. Then $m^2 + 2 \equiv 2 \pmod{4}$, which combined the fact that $4 \mid b^n$ (since $n \geq 2$) and equation (4) result in a contradition. Hence $m$ is odd. Then $a$ is even by (3). If $3 \mid m$, then $a=36$ and $b=1$, yielding $m^2=-1$ by (4). Thus $3 \nmid m$, implying $3 \mid m^2 + 2$. Consequently $3 \mid b^n$ by (4), i.e. $3 \mid b$. This combined with $2 \mid a$ and $a$ and $b$ are coprime natural numbers satisfying $ab = 36 = 2^2 \cdot 3^2$ give us $a=4$ and $b=9$. Hence by (4) we have $(5) \;\; (3^n - m)(3^n + m) = 2$. Therefore $3^n - m = 1$ and $3^n + m = 2$, yielding $(3^n + m) - (3^n - m) = 2 - 1$, i.e. $2m=1$. This contradiction leads us to the following Conclusion: The Diophantine equation (1) has no solutions in natural numbers $m$ and $n$. .

den_thewhitelion wrote: Find n such that $36^n-6$ is the product of three consecutive natural numbers I think that the problem of JBMO 2010 Shortlist is a little different. http://www.docfoc.com/jbmo-shortlist-2010-ahT2a

Original problem : Find all positive integers $n$ such that $36^n-6$ is a product of two or more consecutive positive integers. Let $36^n-6=(k)(k+1)...(k+l-1),k\geq 1,l \geq 2$ If $l=3$ there are not solutions (proved by Solar Plexsus) If $l\geq 4$ than $36^n-6$ is divided by 4 as product 4 or more consecutive integers. But $36^n-6\equiv 2$(mod 4) -contradiction. So $l=2$ $36^n-6=k^2+k$ $4*36^n = (2*6^n)^2=(2k+1)^2+23$ $(2*6^n-2k-1)(2*6^n+2k+1)=23$ $2*6^n-2k-1=1,2*6^n+2k+1=23$ $k=5,n=1$ So answer: $n=1$

Which book is that (because I see it said "Chapter 1")

Thanks @solar plexsus

Answer: No solution. $36^n-6=(x-1)x(x+1)=x^3-x$ \[2 \equiv x^3-x(mod 7)\]Contradiction

Suppose that we had that $(m-1)m(m+1)=36^n-6$ for some pair $(m,n)$ then this can be re-written as $m^3-m+6=6^{2n}$ and then factoring gives $(m+2)(m^2-2m+3)=6^{2n}$ so now notice from simple euclid alg we can get $\gcd(m+2, m^2-2m+3) \mid 11$, but since $11^2 \not \; \mid 6^{2n}$ we must have that $\gcd(m+2, m^2-2m+3)=1$ and thus $m+2=6^a$ and $m^2-2m+3=6^b$ where $a+b=2n$, now we get rid of some base cases, for $n=1$ it trivially fails and now for $n \ge 2$ its clear that $m \ge 3$ and also that therefore $a, b \ge 1$ and thus $m$ is even but also then $4 \mid (m-2)m$ and if $b \ge 2$ then $4 \mid 3$ which can't happen so $b=1$ but then once again $3 \equiv 2 \pmod 4$, can't happen, therefore no such pairs $(m,n)$ exist.

$(x-1)(x-2)(x-3)=36^n-6$ $x(x^2-6x+11)=2^{2n}\cdot 3^{2n}$ $\gcd(x,x^2-6x+11) \in \{1,11\}$ but $11 \not | 36$. Therefore $x$ and $x^2-6x+11$ must be squares. $x^2-6x+11=(x-3)^2+2$ cannot be a square.

Мore generally: The equation $m(m+1)(m+2)=6^n-6$ has a unique solution $(5,3)$ in positive integers. (also $m=0,-1,-2$ in integers)