product of six consecutive integers being a perfect square

Question

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?

Answer 1

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.

Answer 2

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 $x$.

Answer 3

There is a short proof given in an old math journal (The Analyst) by G. W. Hill. You may find it here: https://www.jstor.org/stable/2635974. I replicate it below with annotations.

Let the six consecutive integers be represented as

$\frac{n-5}{2}, \frac{n-3}{2}, \frac{n-1}{2}, \frac{n+1}{2}, \frac{n+3}{2}, \frac{n+5}{2}$ for some odd $n$.

The product of which becomes $\frac{n^2-25}{4} * \frac{n^2-9}{4} * \frac{n^2-1}{4}$

Substitute $x = \frac{n^2-9}{4}$ (an integer as it is the product of two integers) and the product becomes

$x(x+2)(x-4)$

Let $x = k^2y$ such that $k^2$ is the largest square factor contained in $x$. Note $y$ cannot contain a square factor (other than 1), otherwise $k^2$ would not be the largest square factor contained in $x$.

We now examine the equation

$k^2y(k^2y+2)(k^2y-4) = □$ (using old notation where □ is a placeholder for a square integer).

In order for the left hand side to be square, then $y(k^2y+2)(k^2y-4)$ must be square.

$y(k^2y+2)(k^2y-4) = □$

Since $y$ is not a square, then $y$ must be contained as a factor in $(k^2y+2)(k^2y-4)$ and so

$y | (k^2y+2)(k^2y-4) \implies y|8 \implies y = 1 \text{ or } 2$

Case 1: y = 1

$(k^2+2)(k^2-4) = □$

$9 + □ = (k^2-1)^2 \implies k^2-1 = 5 \implies k = \sqrt6$ and hence $k$ is not an integer (contradiction).

Case 2: y = 2

$2*(2k^2+2)(2k^2-4) = □$

Since every square has the form $3n$ or $3n+1$, $k^2$ must be of the form $3n$ or $3n+1$. But, in either case, the resultant form of $□$ becomes $3n+2$, and hence $□$ is not square (contradiction).

Since $y$ cannot equal 1 or 2 then it follows that the product is not square. ■

Notes.

In the paper, the author doesn't explicitly show that $9 + □ = (k^2-1)^2 \implies □ = 16 \text{ and } (k^2-1)^2 = 25 $. However, since we know the side length is 3, we can conclude (3, 4, 5) is the only Pythagorean triple possible. This follows from Euclid's formulation. $3 = m^2 - n^2 = (m-n)(m+n)$. The only factors of 3 are 1 and 3 so m - n = 1 and m + n = 3 and so m = 2 and n = 1. This implies the even side length is 2mn = 4 and the hypotenuse is 5.

Proof that a square must be of the form 3n or 3n+1 is found here.

Answer 4

As mercio says, each number is a square multiplied by multiples of 2, 3 and 5, and no two are multiplied by the same factor. The only possible factors are

5  6 1 2 3 10
10 3 2 1 6  5

That’s three pairs where 2x^2 is very close to y^2, so $x/y \approx \sqrt 2$. There is a well known problem to find squares x^2 so that 2x^2 +/- 1 is a square. The solutions to this times 5 are the only possible values for the first and last number, and they grow exponentially. The same for the two multiples of three. I think we can show that these two sets of numbers don’t fit together.

Answer 5

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.