Perfect powers from products of consecutive terms in arithmetic progression

Powers from Products of Consecutive Terms in Arithmetic Progression

A celebrated theorem of Erdős and Selfridge [14] states that the product of consecutive positive integers is never a perfect power. A more recent and equally appealing result is one of Darmon and Merel [11] who proved an old conjecture of Dénes to the effect that there do not exist three consecutive nth powers in arithmetic progression, provided n 3. One common generalization of these problems ...

Powers from five terms in arithmetic progression

has only the solution (n, k, b, y, l) = (48, 3, 6, 140, 2) in positive integers n, k, b, y and l, where k, l ≥ 2, P (b) ≤ k and P (y) > k. Here, P (m) denotes the greatest prime factor of the integer m (where, for completeness, we write P (±1) = 1 and P (0) = ∞). Rather surprisingly, no similar conclusion is available for the frequently studied generalization of this equation to products of con...

Perfect Powers from Products of Terms in Lucas Sequences

Suppose that {Un}n≥0 is a Lucas sequence, and suppose that l1, . . . , lt are primes. We show that the equation Un1 · · ·Unm = ±l x1 1 · · · l xt t y , p prime, m < p, has only finitely many solutions. Moreover, we explain a practical method of solving these equations. For example, if {Fn}n≥0 is the Fibonacci sequence, then we solve the equation Fn1 · · ·Fnm = 2 x1 · 32 · 53 · · · 541100y under...

Perfect powers in arithmetic progression 1 PERFECT POWERS IN ARITHMETIC PROGRESSION. A NOTE ON THE INHOMOGENEOUS CASE

We show that the abc conjecture implies that the number of terms of any arithmetic progression consisting of almost perfect ”inhomogeneous” powers is bounded, moreover, if the exponents of the powers are all ≥ 4, then the number of such progressions is finite. We derive a similar statement unconditionally, provided that the exponents of the terms in the progression are bounded from above.

Perfect powers in products of terms in an arithmetical progression III