20200321, 21:47  #1 
Mar 2018
17×31 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
17×83 Posts 
Yes.

20200321, 21:54  #3 
Mar 2018
1000001111_{2} Posts 

20200321, 23:00  #4 
"Curtis"
Feb 2005
Riverside, CA
2×3×733 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
2^{2}·1,483 Posts 
Quote:
I don't know of any modular obstructions. 

20200322, 08:39  #6 
"Robert Gerbicz"
Oct 2005
Hungary
17·83 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
2^{2}·1,483 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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