find all $n\in\mathbb{N}$ such that $5^n-1$ can be written as a product of Multiplication of $2q$ consecutive numbers.
because $5^n-1$ cant be divisible by $5$ so $q\in\mathbb{{1,2}}$ and because $4$ divides $5^n-1$ if want to write it in $a(a+1)$ form so $4\mid a$ but I dont know how to continue...
By Pigeonhole Principle, we have $q=1$ or $q=2$.
If $q=1$, we have $k(k+1)=5^n-1$ for some integer $k$. However, the $RHS \equiv 4 \pmod{5}$ for $n\geq 1$ and the $LHS \in \{0,1,2\} \pmod{5}$, so this case is impossible.
If $q=2$, we have $k(k+1)(k+2)(k+3)=5^n-1$ for some integer $k$. Thus, we have $2k(2k+2)(2k+4)(2k+6)=16(5^n-1)$. Letting $x=2k+3$ yields $x^4-10x^2+25=16 \cdot 5^n$. Thus, $x$ is divisible by $5$. Let $x=5y$, so we have $25y^4-10y^2+1=16 \cdot 5^{n-2}$, so we must have $n=2$, from which we get $k=1$.
Therefore, the only solution is $n=2$, and can be written as $1 \times 2 \times 3 \times 4$.