Three consecutive numbers

If a and b are relatively prime, and ab is a square, then a and b are squares. (This is left as an exercise.)

Suppose (n - 1)n(n + 1) = k^2, where n > 1. Then n(n^2 - 1) = k^2. But n and (n^2 - 1) are relatively prime. Therefore n^2 - 1 is a perfect square, which is a contradiction.

Four consecutive numbers
n(n + 1)(n + 2)(n + 3) = (n^2 + 3n + 1)^2 - 1
Five consecutive numbers

Assume the product is a integer square, call it m.

The prime factorization of m must have even numbers of each prime factor.

For each prime factor, p, of m, p >= 5, p^2k must divide one of the

consecutive naturals in the product. (Otherwise, the difference between two of the naturals in the product would be a positive multiple of a prime >= 5. But in this problem, the greatest difference is 4.) So we need only consider the primes 2 and 3.

Each of the consecutive naturals is one of:

1. a perfect square 2. 2 times a perfect square 3. 3 times a perfect square 4. 6 times a perfect square.

By the shoe box principle, two of the five consecutive numbers must fall into the same category.

If there are two perfect squares, then their difference being less than five limits their values to be 1 and 4. (0 is not a natural number, so 0 and 1 and 0 and 4 cannot be the perfect squares.) But 1*2*3*4*5=120!=x*x where x is an integer.

If there are two numbers that are 2 times a perfect square, then their difference being less than five implies that the perfect squares (which are multiplied by 2) are less than 3 apart, and no two natural squares differ by only 1 or 2.

A similar argument holds for two numbers which are 3 times a perfect square.

We cannot have the case that two of the 5 consecutive numbers are multiples (much less square multiples) of 6, since their difference would be >= 6, and our span of five consecutive numbers is only 4.

Therefore the assumption that m is a perfect square does not hold.


In general the equation
y^2 = x(x+1)(x+2)...(x+n), n > 3

has only the solution corresponding to y = 0.

This is a theorem of O. Rigge, "Uber ein diophantisches Problem," IX Skan. Math. Kong. Helsingfors (1938) and P. Erdos, "Note on products of consecutive integers," J. London Math. Soc. #14 (1939), pages 194-198.

A proof can be found on page 276 of L. Mordell, "Diophantine Equations", Academic Press 1969.

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