olymmath

BG-MO 2012 Problem 1

    The sequence a1,a2,a3…a_1,a_2,a_3\ldots , consisting of natural numbers, is defined by the rule: an+1=an+2t(n)a_{n+1}=a_{n}+2t(n) for every natural number nn , where t(n)t(n) is the number of the different divisors of nn (including 11 and nn ). Is it possible that two consecutive members of the sequence are squares of natural numbers?

    Dataset text AI-MO AoPS. Quoted with source.

    Solutions

    No solution is on olymmath yet. Official and human solutions come first when they arrive. AI solutions come after them, with a label.

    Add a solution

    Similar problems

    These problems have the most words in common with this statement.

    Locator bg-mo/2012/p1 · Tag VCSDFTF700 · Edit this page on GitHub · Keys: j next, k previous