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Part of 2013 IMAC Arhimede

Problems(1)

4^{6^n}+1943 dividible by 2013

Source: IMAC Arhimede 2013 p2

5/6/2019
For all positive integer nn, we consider the number an=46n+1943a_n =4^{6^n}+1943 Prove that ana_n is dividible by 20132013 for all n≥1n\ge 1, and find all values of nn for which an−207a_n - 207 is the cube of a positive integer.
number theoryperfect cubedivisibleexponential