Show that if an (infinite) arithmetic progression of positive integers contains an nth power, then it contains infinitely many nth powers.
Solution
Let the difference between adjacent terms of the progression be d. Suppose that Nn is an nth power in the progression. Then (N+d)n is a larger nth power in the progression (it is obvious from the binomial expansion that it has the form N + kd).
© John Scholes
jscholes@kalva.demon.co.uk
25 Jan 2002