# product of six consecutive integers being a perfect square

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?

-
I do not have so much experience to prove things, but I would start by the following steps: $k(k + 1) = u$, so the square root of $u$ would be the square of $k$ multiplied by $k + 1$, and this makes us think of two consecutive numbers that satisfy that condition. Extend this for 6 numbers, and this seems to be hard to think in numbers that can satisfy that. –  Ian Mateus Dec 12 '11 at 19:18
The square root of $k$ multiplied by the square root of $k + 1$ - sorry. –  Ian Mateus Dec 12 '11 at 19:25
You're referring to "Narumi, S.: An extension of a theorem of Liouville’s. Tohoku Math. J. 11, 128–142 (1917)". I doubt if you'll find it online ; it should be available in your library. –  Jacob Dec 12 '11 at 19:28
@P23 The paper by Narumi is available here. –  t.b. Dec 12 '11 at 20:12

For the record, here is a very simple proof that the product of six consecutive positive integers cannot be a perfect square. Let $n_1,n_5$ be the two of the six that are congruent modulo $6$ to $1$ and $5$ respectively. Then $|n_1-n_5|\in\{2,4\}$ and any prime factor of $n_1$ or of $n_5$ must be at least $5$. Given their distance, such a prime factor cannot divide the other of these two numbers, and in order to make the product a perfect square it must have even multiplicity in the number that it does divide. But then (by the fundamental theorem of arithmetic if you like) $n_1$ and $n_5$ are themselves both (nonzero) perfect squares, which is incompatible with $|n_1-n_5|\in\{2,4\}$.

-
That's a nice proof. –  Soham Chowdhury Sep 7 '13 at 6:14

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.

-
That's very nice. But I suppose the definition of "elementary" I originally had in mind would not involve knowing anything about elliptic curves. As mentioned, the desired "quick" and "elementary" proof may or may not exist. –  idmercer Dec 12 '11 at 23:44

It's possible to adapt Ross Millikan's argument :

Assuming none of the numbers is $0$, since primes greater than $5$ can only appear once, each of the six numbers is of the form $2^a 3^b 5^c y^2$ with $a,b,c = 0$ or $1$. Also, each prime $2,3,5$ can only appear an even number of time in order for the product to be a square.

If two of the six numbers have the same exponent triple $(a,b,c)$, it puts a very small upper bound on $x$ because it implies that we have two squares $y^2$ extremely close together. So the goal is to try to give them six different exponent triples out of the $8$ available and fail.

If the prime $5$ doesn't appear, you only have $4$ triples for $6$ numbers, impossible. So $5$ has to appear twice in $x$ and $x+5$ only.

Then the four numbers in the middle have to take the other $4$ triples with no $5$, so the prime $3$ must appear in $x+1$ and $x+4$ only.

Finally, the prime $2$ has to appear twice in the middle numbers only. It can't appear three times or else the whole product wouldn't be a square, and it can't appear four times because there is not enough room.

Thus the numbers $x$ and $x+5$ must be of the form $5y^2$. So whatever we do we get an upper bound on $x$.

-
Very good. I was playing around with that type of reasoning but for some silly reason missed following it to its conclusion. I'll wait a little longer just in case there's an even faster "trick" we all missed, but this might be essentially "the" elementary way to prove it. It's like the argument for three or five consecutive integers, but maybe there's no way around the fact that the details for six consecutive integers are a tiny bit more messy. –  idmercer Dec 13 '11 at 1:01
@idmercer: Dear mercio and idmercer, Note that the Erdos's argument begins in the same way. He then replaces the explicit consideration of "triples" with more general considerations of the distribution of square free numbers. However, his argument does have the same flavour as this one, just generalized to the arbitrary length case. Regards, –  Matt E Dec 13 '11 at 3:04

For two integers, note that the GCD of $k$ and $k+1$ is $1$, so any number that divides $k$ does not divide $k+1$. Then if $n^2=k(k+1)$, both $k$ and $k+1$ would have to be squares, but the no squares of positive integers differ by $1$.

Added: as idmercer suggests $k(k+1)(k+2)(k+3)=(k^2+3k+1)^2-1$ so cannot be a square.

-
Or you can just note that $k^2<k(k+1)<(k+1)^2$ :) –  Thomas Andrews Dec 12 '11 at 21:03
There's also a very short proof that k(k+1)(k+2)(k+3) can't be a perfect square. –  idmercer Dec 12 '11 at 23:45
The basis of this argument is finding a prime number $p>5$ that divides exactly one number of the list. However, the argument looks over the possibility that $p$ divides that number twice (or more), i.e., that $p^2$ divides that number; this invalidates the argument. To see how one can repair this problem, see my answer. –  Marc van Leeuwen Sep 7 '13 at 6:48