20200321, 21:47  #1 
Mar 2018
531_{10} Posts 
Diophantine equation
Are there infinitely many solutions to these Diophantine equation
10^na^3b^3=c^2 with n, a, b, c positive integers? 
20200321, 21:51  #2 
"Robert Gerbicz"
Oct 2005
Hungary
1,531 Posts 
Yes.

20200321, 21:54  #3 
Mar 2018
3^{2}·59 Posts 

20200321, 23:00  #4 
"Curtis"
Feb 2005
Riverside, CA
2·3·859 Posts 
Take an algebra class. Learn how to answer your own questions.
Even wikipedia articles would give you sufficient tools to answer your curiosities. 
20200322, 04:46  #5  
Aug 2006
3·1,993 Posts 
Quote:
I don't know of any modular obstructions. 

20200322, 08:39  #6 
"Robert Gerbicz"
Oct 2005
Hungary
1,531 Posts 
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.

20200323, 03:03  #7  
Aug 2006
3·1,993 Posts 
Quote:
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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