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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?

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Brian, I see that this is your first question. So I wanted to let you know a few things about MathSE. We like to know the sources of questions. We also like to know what you've tried on a problem. These sort of pleasantries usually result in more and better answers. Finally, I should add that posting questions in the imperative (i.e. Compute all such...) is considered rude by some of the members, so I advise you to change that wording. – mixedmath Aug 28 '11 at 0:02
@mixedmath: are you sure this is not $5^z - 1$? – Henry Aug 28 '11 at 1:02
Note that 6 or more consecutive positive integers must include a multiple of 5, and so there can be no solutions with 6 or more positive integers. Combine that with Byron Schmuland's solution of the case of 4 consecutive positive integers and user14044's disposition of the case of 2 consecutive positive integers, and you are done. – tzs Aug 28 '11 at 20:09

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 $$(-1)(-2)(-3)(-4)=(4)(3)(2)(1)=5^2-1.$$

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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).

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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.


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.

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Technically, zero is an even number... – Ilmari Karonen Aug 28 '11 at 0:25
I would have said $5^{z-1}$ is an odd number, but this amount to something similar. – Henry Aug 28 '11 at 1:01

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