A236211 Numbers c > 0 for which there exist integers a > 1 and b > 1 such that the equation a^x - b^y = c has two solutions in positive integers x, y.

1, 3, 4, 5, 9, 10, 13, 89, 275, 1215, 4900

Offset

1,2

Comments

Bennett proved that if a, b, c are nonzero integers with a > 1 and b > 1, then the equation a^x - b^y = c has at most two solutions in positive integers x and y.

Bennett conjectured that if a, b, c are positive integers with a > 1 and b > 1, then the equation a^x - b^y = c has at most one solution in positive integers x and y, except for the triples (a,b,c) = (3,2,1), (2,5,3), (6,2,4), (2,3,5), (15,6,9), (13,3,10), (2,3,13), (91,2,89), (280,5,275), (6,3,1215), (4930,30,4900). If this is true, then the present sequence is complete.

References

R. K. Guy, Unsolved Problems in Number Theory, D9.

T. N. Shorey and R. Tijdeman, Exponential Diophantine Equations, Cambridge University Press, 1986.

Links

Table of n, a(n) for n=1..11.

M. A. Bennett, On Some Exponential Equations of S. S. Pillai, Canad. J. Math., 53 (2001), 897-922.

J.-H. Evertse, Review of M. A. Bennett's "On Some Exponential Equations of S. S. Pillai", zbMATH 0984.11014

OEIS, Entries related to Pillai's equation

M. Waldschmidt, Open Diophantine problems

E. Weisstein's MathWorld, Pillai's Conjecture

Example

3 - 2 = 3^2 - 2^3 = 1.
2^3 - 5 = 2^7 - 5^3 = 3.
6 - 2 = 6^2 - 2^5 = 4.
2^3 - 3 = 2^5 - 3^3 = 5.
15 - 6 = 15^2 - 6^3 = 9.
13 - 3 = 13^3 - 3^7 = 10.
2^4 - 3 = 2^8 - 3^5 = 13.
91 - 2 = 91^2 - 2^13 = 89.
280 - 5 = 280^2 - 5^7 = 275.
6^4 - 3^4 = 6^5 - 3^8 = 1215.
4930 - 30 = 4930^2 - 30^5 = 4900.

Crossrefs

Cf. A207079 and the OEIS link.

Sequence in context: A047249 A327257 A228941 * A279616 A050068 A250444

Adjacent sequences: A236208 A236209 A236210 * A236212 A236213 A236214

Keyword

nonn

Author

Jonathan Sondow, Jan 23 2014

Status

approved