Diophantine equation
Are there infinitely many solutions to these Diophantine equation
10^na^3b^3=c^2 with n, a, b, c positive integers? 
Yes.

[QUOTE=R. Gerbicz;540413]Yes.[/QUOTE]
How to proof it? 
Take an algebra class. Learn how to answer your own questions.
Even wikipedia articles would give you sufficient tools to answer your curiosities. 
[QUOTE=enzocreti;540412]Are there infinitely many solutions to these Diophantine equation
10^na^3b^3=c^2 with n, a, b, c positive integers?[/QUOTE] It seems rather difficult to find positive integers n such that there are no positive integers a, b, and c with 10^n = a^3 + b^3 + c^2. n = 5 is an example, but there are no others up to 18. I don't know of any modular obstructions. 
[QUOTE=CRGreathouse;540436]It seems rather difficult to find positive integers n such that there are no positive integers a, b, and c with 10^n = a^3 + b^3 + c^2. n = 5 is an example, but there are no others up to 18.
I don't know of any modular obstructions.[/QUOTE] Multiple the equation by 10^(6*k), since it is cube and square if there is a solution for n=N, then there is a solution for N+6*k. 
[QUOTE=R. Gerbicz;540442]Multiple the equation by 10^(6*k), since it is cube and square if there is a solution for n=N, then there is a solution for N+6*k.[/QUOTE]
Ah! of course. So it suffices to show 10^1 = 1^3 + 2^3 + 1^2 10^2 = 3^3 + 4^3 + 3^2 10^3 = 6^3 + 7^3 + 21^2 10^4 = 4^3 + 15^3 + 81^2 10^6 = 7^3 + 26^3 + 991^2 10^11 = 234^3 + 418^3 + 316092^2 and then we know that all powers of 10, other than 10^5, are expressible as the sum of two positive cubes and a positive square. 
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