If a and b are relatively prime, and ab is a square, then a and b are squares. (This is left as an exercise.)
- n(n + 1)(n + 2)(n + 3) = (n^2 + 3n + 1)^2 - 1
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.
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.