# How surprising is this result on the diophantine equation $x(x+1)…(x+n-1) = y^n+k$?

A number of years ago, I proved the following result:

For any integer $k$, the number of positive integral solutions to $x(x+1)...(x+n-1) = y^n+k$ with $n \ge 3$ is finite (i.e., there are only a finite number of $(x, y, n)$ satisfying this equation for any $k$).

It is pretty clear that for any fixed $k$ and $n$ there are only a finite number of $x$ and $y$ (you can prove that $y \le |k|$), but the fact that there are only a finite number of $n$ came as a surprise to me. I initially proved that $n < e|k|$ and later derived much stricter bounds.

The way I phrased this is "The product of $n$ consecutive integers is almost never close to an $n$-th power."

My question is whether this result is surprising?

Thanks.

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Surprise is relative. Perhaps you might like to ask a more precise question, like "are there any common heuristics which suggest that this is the obvious answer" or "are there elementary techniques which quickly yield this result"? – Qiaochu Yuan Aug 27 '11 at 3:00
@ Qiaochu Yuan: I seem to recall a general result that when the sum of degrees of a Diophantine equation (here 2/n) is less than 1 you should expect a finite number of solutions, but a quick search doesn't turn it up. Maybe this would answer the question. – Ross Millikan Aug 27 '11 at 3:22
@Ross: this should follow from Scott Carnahan's answer in the linked MO post, but it only gives the result for fixed $n$. – Qiaochu Yuan Aug 27 '11 at 3:26

The result is not surprising in the sense that standard probabilistic heuristics suggest it to be true. In the interval $[2^d, 2^{d+1})$ there are $2^d$ numbers. Of those numbers, $O(d \cdot 2^{d/3})$ are perfect third powers or higher, so the probability that a number in this interval is a perfect third power or higher is $O(d \cdot 2^{-2d/3})$. The probability that a number in this interval is of the form $x(x+1)...(x+n-1) - k$ for some $n \ge 3$ should be approximately the same, and if we assume that the two events are independent, then the probability that a solution exists to your Diophantine equation with $y^n \in [2^d, 2^{d+1})$ is $O(d^2 \cdot 2^{-4d/3})$.

The sum of these probabilities over all $d$ converges. A heuristic application of Borel-Cantelli then suggests that the number of solutions is finite.

Of course, this argument also applies, incorrectly, to $n = 2$. Here there is algebraic structure which violates the independence assumption. But this structure doesn't generalize to higher $n$, and I guess this can be made precise by Faltings' theorem for fixed $n$.

Speaking of which, Scott Carnahan's answer to this MO question about the Mordell conjecture may be more convincing to you, as it correctly excludes the case $n = 2$. (Take the sum over all $d \ge 3$, in the notation of his answer.)

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BTW, here is how I converted the bound on $y$ to a bound on $n$:

Since $2(n/e)^n < n!$,

$$2(n/e)^n < n! \le x(x+1)...(x+n-1) = y^n+k.$$

Using the corrected bound on $y$ of $y \le |k|$,

$$2(n/e)^n < |k|^n + |k| \le 2|k|^n$$

so $n/e < |k|$ or $n < e|k|$.

I did this over 40 years ago, but I still remember thinking that this was magic, somehow pulling the result out of thin air.

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Nothing particularly magical seems to be going on here given that $y \le |k|$: you're just using the fact that factorials grow faster than exponentials. What I'd really like to know is how you showed that $y \le |k|$. – Qiaochu Yuan Aug 27 '11 at 4:49
I agree with Quaochu Yuan. – mick Dec 5 '13 at 22:55

It is a result of Erdos and Selfridge that the product of consecutive integers is never a power (with trivial exceptions). Perhaps something in their paper is of use here.

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