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 2020-03-21, 21:47 #1 enzocreti   Mar 2018 32·59 Posts Diophantine equation Are there infinitely many solutions to these Diophantine equation 10^n-a^3-b^3=c^2 with n, a, b, c positive integers?
 2020-03-21, 21:51 #2 R. Gerbicz     "Robert Gerbicz" Oct 2005 Hungary 27738 Posts Yes.
2020-03-21, 21:54   #3
enzocreti

Mar 2018

32·59 Posts

Quote:
 Originally Posted by R. Gerbicz Yes.
How to proof it?

 2020-03-21, 23:00 #4 VBCurtis     "Curtis" Feb 2005 Riverside, CA 7·11·67 Posts Take an algebra class. Learn how to answer your own questions. Even wikipedia articles would give you sufficient tools to answer your curiosities.
2020-03-22, 04:46   #5
CRGreathouse

Aug 2006

175B16 Posts

Quote:
 Originally Posted by enzocreti Are there infinitely many solutions to these Diophantine equation 10^n-a^3-b^3=c^2 with n, a, b, c positive integers?
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.

2020-03-22, 08:39   #6
R. Gerbicz

"Robert Gerbicz"
Oct 2005
Hungary

1,531 Posts

Quote:
 Originally Posted by CRGreathouse 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.
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.

2020-03-23, 03:03   #7
CRGreathouse

Aug 2006

3·1,993 Posts

Quote:
 Originally Posted by R. Gerbicz 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.
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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