Product of an even number of consecutive positive integers Let $z$ be a positive integer. How should one compute all $z$ such that $5^z-1$ can be written as the product of an even number of consecutive positive integers?
 A: Here is the case of four consecutive integers. Suppose that the integer $x$ solves 
$x(x-1)(x-2)(x-3)=5^z-1$ with $z$ a positive integer. Then we have $x(x-1)(x-2)(x-3)+1=5^z$, where the
left hand side can be factored as $(x^2-3x+1)^2$. Thus $x^2-3x+1$ must also be
a positive power of $5$. 
Working modulo $5$, we have 
$$0\equiv x^2-3x+1\equiv x^2+2x+1\equiv (x+1)^2,$$
so that $x\equiv -1\pmod 5$.
Thus we can write $x=5q-1$ for some integer $q$.
Substituting this back into $x^2-3x+1$ gives 
$25q(q-1)+5$ which can only be a power of $5$ if $q=0$ or $q=1$.
Thus $x$ must be  either $-1$ or $4$, and this gives  two solutions
$$(-1)(-2)(-3)(-4)=(4)(3)(2)(1)=5^2-1.$$
A: The case for two integers can be discounted easily... note that $5^z -1$ is congruent to $4 \pmod 5$, whereas $a(a+1)$ is never congruent to $4 \pmod 5$ (easily checked).
A: This question is sort of funny to me. By the Fundamental Theorem of Arithmetic, any integer can be uniquely written as a product of primes. For this particular question, if we call $n := 5^{z - 1}$, then we know exactly the prime decomposition of n. In particular, only one prime divides n. Thus 2 does not divide n, ever. As every other number is even, no even number of consecutive positive integers will ever divide this number.
ADDED
The question is actually about $5^z - 1$, not $5^{z-1}$. So my answer above is incorrect, but I keep it for posterity. I will say a big hint - we only care about the product of 2 or 4 consecutive positive integers, as 5 does not divide $5^z - 1$ but 5 will always divide products of 5 or more integers. This vastly simplifies things.
