Prove that there are no perfect cubes in the infinite set {83, 8383, 838383 ...}?

Question

First, work with the subset {83, 838, 838383 ...}:
Assume that there exists a number x for which x^3 = 8383...83
If x^3 terminates in a 3, then x must terminate in a 7 (--7^3 = 343).
x can be rewritten as (10n+7):
(10n+7)^3 = 8383...83
1000n^3+1400n^2+490n+700n^2+980N+343 = 8383...83
1000n^3+2100n^2+1470n+343 = 8383...83
-subtract 343 on both sides
1000n^3+2100n^2+1470n = 8383...838040
-divide by 10 on both sides
100n^3+210n^2+147n = 8383...83804

Since the value of n must make 100n^3+210n^2+147n a number terminating in 4, and 147n is the only part of the equation that concerned the ones digit, 147n must equal a number terminating in 4. In order for 147n to terminate in a 4, n must terminate in a 2.
Therefore n can be rewritten as (10m+2)
(10(10m+2)+7)^3 = 8383...83
(100m+20+7)^3 = 8383...83
(100m+27)^3 = 8383...83
100000m^3+81000m^2+14310m+4580=8383...8383
-subtract 4580 on both sides
100000m^3+81000m^2+14310m= 8383...833803

There is a contradiction here; while the left side of the equation is a multiple of 10, the right side of the equation is not a multiple of 10. Therefore, the original assumption that there exists a number x for which x^3=8383...83 is false and there are no perfect cubes in the set {83, 8383, 838383 ...}

Answer 1

I'm not sure what your QUESTION is, but it sure seems like you're on a very good track with that.