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
2·7·103 Posts 
Yes.

20200321, 21:54  #3 
Mar 2018
17×31 Posts 

20200321, 23:00  #4 
"Curtis"
Feb 2005
Riverside, CA
4,673 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
1754_{16} Posts 
Quote:
I don't know of any modular obstructions. 

20200322, 08:39  #6 
"Robert Gerbicz"
Oct 2005
Hungary
2·7·103 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}×1,493 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. 

Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Basic Number Theory 13: Pythagorean triples and a few Diophantine equations  Nick  Number Theory Discussion Group  2  20161218 14:49 
Diophantine Equation  flouran  Math  7  20091212 18:48 
Diophantine Question  grandpascorpion  Math  11  20090923 03:30 
Simple Diophantine equation  Batalov  Puzzles  3  20090410 11:32 
Diophantine problem  philmoore  Puzzles  8  20080519 14:17 