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Old 2020-03-23, 03:03   #7
CRGreathouse
 
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Aug 2006

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Cool

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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