Products Of Consecutive Integers Research Articles

How Many Pairs of Products of Consecutive Integers Have the Same Prime Factors?

How Many Pairs of Products of Consecutive Integers Have the Same Prime Factors?

The product of consecutive integers is never a power

has no solution in integers with k >_ 2, 1 >_ 2 and n >_ 0 . (These restrictions on k, 1 and n will be implicit throughout this paper .) The early literature on this subject can be found in Dickson's history and the somewhat later literature in the paper of Oblath [5] . Rigge [6], and a few months later Erdos [1], proved the conjecture for 1 = 2 . Later these two authors [1] proved that for fixed 1 there are at most finitely many solutions to (1) . In 1940, Erdos and Siegel jointly proved that there is an absolute constant c such that (1) has no solutions with k > c, but this proof was never published . Later Erdos [2] found a different proof ; by improving the method used, we can now completely establish the old conjecture . Thus we shall prove

On Equal Products of Consecutive Integers

Using the theory of algebraic numbers, Mordell [1] has shown that the Diophantine equation1possesses only two solutions in positive integers; these are given by n = 2, m = 1, and n = 14, m = 5. We are interested in positive integer solutions to the generalized equation2and in this paper we prove for several choices of k and l that (2) has no solutions, in other cases the only solutions are given, and numerical evidence for all values of k and l for which max (k, l) ≤ 15 is also exhibited.

On the product of consecutive integers. III

On the product of consecutive integers. III

Note on the Product of Consecutive Integers (II)

Note on the Product of Consecutive Integers (II)

Note on Products of Consecutive Integers

Note on Products of Consecutive Integers