# Product of Four Consecutive Integers

Product of four consecutive positive integers cannot be a square. Moreover, it is always 1 less than a square of an integer. Why? I was unaware of the popularity of this question until informed by José Antônio Fabiano Mendes of Rio de Janeiro, Brazil. He also suggested the following solution:

Let Q = n(n + 1)(n + 2)(n + 3).  Observe that (n + 1)(n + 2) - n(n + 3) = 2  and define P = (n + 1)(n + 2) - 1  so that (n + 1)(n + 2) = P + 1  and n(n + 3) = P - 1,  implying that Q = P^2 - 1.

As it came out, several proofs of this result are known. For example, one can check directly that

 n(n + 1)(n + 2)(n + 3) + 1 = (n^2 + n)(n^2 + 5n + 6) + 1 = n^4 + 6n^3 + 11n^2 + 6n + 1 = (n^2 + 3n + 1)^2.

More interestingly, the fact was investigated by numerous mathematicians, Euler in particular (Dickson, p. 635). In 1780, Euler obtained a theorem one of whose corollaries was a generalization of the question in hand: a product of four consecutive terms of an arithmetic progression can never be a square. (This and other related questions are discussed in an article by Fogarty & O'Sullivan.)

The most general result was obtained by Erdös and Selfridge, viz., the product of consecutive integers is never a power. In other words, the equation

m(m + n)(m + 2n) \times ... \times (m + kn) = r^w

has no integer solutions.

### References

1. L. E. Dickson, History of the Theory of Numbers, v II, Dover, 2005
2. P. Erdös and J. L. Selfridge, The product of consecutive integers is never a power, Illinois J. Math. 19 (1975), 292–301.
3. K. Fogarty, C. O’Sullivan, Arithmetic Progressions with Three Parts in Prescribed Ratio and a Challenge of Fermat