A 1939 paper of Erdos (Note on Products of Consecutive Integers, J. London Math. Soc. 14 (1939), 194–198) shows that a product of consecutive positive integers cannot be a perfect square. He cites a 1917 paper by Narumi which proves that a product of at most 202 consecutive positive integers cannot be a perfect square. I cannot seem to easily find Narumi's paper.
Although this result is known, I am curious about self-contained elementary proofs of special cases.
It's not too difficult to come up with fairly quick proofs for two, three, four, five, or seven consecutive integers.
Is there a short self-contained elementary proof that the product of six consecutive positive integers cannot be a perfect square? Or is it perhaps fair to say that this is the first "tricky" case?
Best Answer
I am not sure how elementary you want your proof to be, but here is a proof that uses elliptic curves...
Suppose that there are $x,y\in\mathbb{Z}$ such that $x>0$ and
$$y^2=x(x+1)(x+2)(x+3)(x+4)(x+5).$$
If we put $t=x+2+\frac{1}{2}$, then we have
$$y^2=(t-5/2)(t-3/2)(t-1/2)(t+1/2)(t+3/2)(t+5/2)=\left(t^2-\frac{1}{4}\right)\left(t^2-\frac{9}{4}\right)\left(t^2-\frac{25}{4}\right),$$
or, equivalently,
$$4^3y^2 = (4t^2-1)(4t^2-9)(4t^2-25).$$
If we put $U=2^3y$ and $V=4t^2$, then we have a solution for the equation
$$U^2=(V-1)(V-9)(V-25)=V^3 - 35V^2 + 259V - 225.$$
This defines an elliptic curve $E/\mathbb{Q}$, and we can use standard techniques to calculate the rank of the group of rational points $E(\mathbb{Q})$. This method ($2$-descent) shows that the rank of the curve is $0$, and one can easily separately show that the torsion subgroup is $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$. It follows that the only points on $E(\mathbb{Q})$ are the trivial points $(V,U)=(1,0)$, $(9,0)$ and $(25,0)$, plus the point ``at infinity'' on the curve. These correspond to $t$-values $t=\pm 1/2$, $\pm 3/2$ and $\pm 5/2$, and therefore do not give any integer values of $x$ with $x> 0$. Hence, there are no integer solutions to our original equation.
There is probably some elementary argument that shows that $U^2=(V-1)(V-9)(V-25)$ only has $3$ solutions, but I can't think of one right away.