The South African Mathematical Olympiad Third Round

$5^n-1$ cannot be a product of more than five consecutive integers: since one of the factors would have to be divisible by $5$, this would have to be the case for the product as well, but $5^n-1$ is not divisible by $5$.

Now we have to consider two cases:

Case 1: If $5^n-1$ is a product of two consecutive integers, say $m$ and $m+1$, then $$5^n=m(m+1)+1=m^2+m+1.$$ Consider this equation modulo $5$: clearly, the left hand side is divisible by $5$. On the other hand, the right hand side is never divisible by $5$:

$m\mathrm{mod}5$	0	1	2	3	4
$m^2+m+1\mathrm{mod}5$	1	3	2	3	1

It follows that there is no solution in this case.

Case 2: If $5^n-1$ is a product of four consecutive integers, say $m,\dots ,m+3$, then $$\begin{array}{rl}{5}^n& =m(m+1)(m+2)(m+3)+1=(m^2+3m)(m^2+3m+2)+1\\ & =(m^2+3m+1{)}^2-1+1\\ & =(m^2+3m+1{)}^2.\end{array}$$ Therefore $n$ must be even, and $m^2+3m+1$ must be a power of $5$: $$m^2+3m+1=5^k,$$ where $k=\frac{n}{2}>0$. Solving for $m$ yields $$m=\frac{-3\pm \sqrt{5+4\cdot 5^k}}{2}=\frac{-3\pm \sqrt{5(1+4\cdot 5^{k-1})}}{2}.$$ If now $k>1$, then $5(1+4\cdot 5^{k-1})$ is divisible by $5$, but not by $25$ ($4\cdot 5^{k-1}$ is a multiple of $5$ in this case, so $4\cdot 5^{k-1}+1$ is not), and so it cannot be a square. This shows that there cannot be any solutions for $k>1$. On the other hand, $k=1$ and $m=1$ satisfy the equation, which leads us to the only solution $$5^2-1=1\cdot 2\cdot 3\cdot 4.$$