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Old 2020-03-21, 21:47   #1
enzocreti
 
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Default 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?
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Old 2020-03-21, 21:51   #2
R. Gerbicz
 
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"Robert Gerbicz"
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Yes.
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Old 2020-03-21, 21:54   #3
enzocreti
 
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Yes.
How to proof it?
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Old 2020-03-21, 23:00   #4
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Take an algebra class. Learn how to answer your own questions.
Even wikipedia articles would give you sufficient tools to answer your curiosities.
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Old 2020-03-22, 04:46   #5
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Quote:
Originally Posted by enzocreti View Post
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.
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Old 2020-03-22, 08:39   #6
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Quote:
Originally Posted by CRGreathouse View Post
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.
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Old 2020-03-23, 03:03   #7
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Quote:
Originally Posted by R. Gerbicz View Post
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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