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?

  • $\begingroup$ 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. $\endgroup$
    – Ian Mateus
    Dec 12, 2011 at 19:18
  • $\begingroup$ The square root of $k$ multiplied by the square root of $k + 1$ - sorry. $\endgroup$
    – Ian Mateus
    Dec 12, 2011 at 19:25
  • $\begingroup$ 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. $\endgroup$
    – Jacob
    Dec 12, 2011 at 19:28
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    $\begingroup$ @P23 The paper by Narumi is available here. $\endgroup$
    – t.b.
    Dec 12, 2011 at 20:12
  • $\begingroup$ Please unaccept my wrong answer, so that I can delete it. $\endgroup$ Dec 1, 2015 at 10:59

6 Answers 6


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


If we put $t=x+2+\frac{1}{2}$, then we have


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.

  • $\begingroup$ 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. $\endgroup$
    – idmercer
    Dec 12, 2011 at 23:44
  • $\begingroup$ Note that with a linear change of variable $\, x = (V-13)/4, y = U/8\,$ the equation $U^2 = (V-1)(V-9)(V-25) $ becomes $y^2 = x^3 + x^3 - 9x - 9$ which is the Elliptic curve with Cremona label 192b2 which confirms that the Mordell-Weil group structure is the Klein four-group with three rational points. These points are integer points also. $\endgroup$
    – Somos
    Nov 29 at 17:36

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 $x$.

  • $\begingroup$ 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. $\endgroup$
    – idmercer
    Dec 13, 2011 at 1:01
  • $\begingroup$ @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, $\endgroup$
    – Matt E
    Dec 13, 2011 at 3:04
  • $\begingroup$ I think I understand what is going on, but the formulation is not easy to follow at some points. You might say somewhere that $x$ is the least of the six consecutive numbers considered. But the most confusing point is that in the beginning "a prime appears" means it divides one of the six numbers, while later it seems to mean occur with odd multiplicity in one of the six numbers. $\endgroup$ Dec 1, 2015 at 12:42
  • $\begingroup$ Either x and x+5 are divisible by 5, or 5^3, or 5^5 etc. or one of the others is divisible by 25. $\endgroup$
    – gnasher729
    Nov 29 at 20:32

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.

  • 4
    $\begingroup$ Or you can just note that $k^2<k(k+1)<(k+1)^2$ :) $\endgroup$ Dec 12, 2011 at 21:03
  • $\begingroup$ There's also a very short proof that k(k+1)(k+2)(k+3) can't be a perfect square. $\endgroup$
    – idmercer
    Dec 12, 2011 at 23:45

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.


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


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. ■


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.

  • 2
    $\begingroup$ While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review $\endgroup$ Nov 29 at 16:40
  • 1
    $\begingroup$ While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review $\endgroup$
    – amWhy
    Nov 29 at 17:00

There's an elementary answer actually. You have n, n+1, n+2, n+3, n+4, n+5. Every two numbers there will be a multiple of two. This means, because there are six numbers in all, that either the first xor the last number will be a multiple of two. There will be three multiples of two. Every three numbers will be a multiple of three. Because three is odd, if the first multiple of three is even, the second will be three plus that and so will be odd. If the first multiple of three is odd, the second is three plus an odd number so is even. So no matter what, one of the multiples of two is also a multiple of three, and there are only two multiples of three in our list. There will be one or two multiples of five in the list. If there are two, then it must be the first and last numbers. Since one of the first and last numbers is already a multiple of two, we only gain information about one "new" number, unless the first is a multiple of two and the last is a multiple of three, or vice versa, in which case we gain information about no "new" numbers. Now if the five appears once in the list only we then gain information about one "new" number at max. So, we have found all the places a 2, 3, or 5 may be a factor -- but this only accounts for five unique numbers (three multiples of 2, one extra multiple of three [because one overlaps with the multiples of two], and one (or none) extra multiple of 5 [because if there are two fives one overlaps with the multiples of two]. That means we have five numbers which have prime factors 2, 3, or 5 in our list, and one which does not. But a list of 6 consecutive positive integers can not repeat any prime factor more than 5. So our unknown number must have a prime factor that is only once in the list, and thus the product of the list cannot be a square. The only exception is the number without any prime factors, 1. If our unknown number is 1, then the list will be 1, 2, 3, 4, 5, 6. But five is a prime factor only once there in that list, so that isn't a square either.

Edit: This is wrong because the number without factors 2, 3, or 5 could be a square itself.

  • 2
    $\begingroup$ A wall of text is never elementary. $\endgroup$ Sep 7, 2013 at 6:16
  • $\begingroup$ Well, I obviously didn't answer very neatly. The question asker's comment to the top answer seems to seek a quicker trick, which this is, though it's presented poorly. The asker seems to be under a misconception that there is no such trick in this case. It's an old question, so maybe he'll never see it, but I got here via google search so it may be of some use to future viewers, so that they aren't misled by the thought that there isn't a trick. If you want to present this answer properly, feel free, I'm certainly not after any points -- I'm just a guest. $\endgroup$
    – Guest
    Sep 7, 2013 at 6:42
  • $\begingroup$ 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. $\endgroup$ Sep 7, 2013 at 6:48
  • $\begingroup$ Right, thanks for for pointing it out. I, too, caught it as I was trying to write the answer out neatly. I guess there are some benefits to a clear presentation after all. $\endgroup$
    – Guest
    Sep 7, 2013 at 7:05
  • $\begingroup$ Who would dare to doubt the benefits of writing out an answer neatly? Which is not to say these benefits come without a lot of effort. The benefits of reading other answers however come at a lesser price. $\endgroup$ Sep 7, 2013 at 7:29

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