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Number Theory Problem in 2012 CMO (created on Wolfram Cloud)

Posted 10 years ago
POSTED BY: Shenghui Yang
4 Replies

enter image description here - another post of yours has been selected for the Staff Picks group, congratulations !

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POSTED BY: EDITORIAL BOARD

Great explanation! How about finding integer solutions for a, b, and c that satisfy:

a^3 + b^3 + c^3 == 33
POSTED BY: Sander Huisman

The moment I see your question my gut feeling tells me there must be no trivial solution. I searched on the Mathworld page of

Diophantine Equation -- 3rd Powers

and found the reference paper

J.C.P. Miller & M.F.C. Woollett, Solutions of the Diophantine equation x3 + y3 + z3 = k, J. London Math. Soc. 30 (1955), 101–110.

There are two other articles reference the same source.

  • Rowland, Known Families of Integer Solutions of x3 + y3 + z3 = n
  • Heath-Brown, On Solving The Diophantine Equation X3 +Y3 +Z3 = K On A Vector Computer

In your case, there is no known solution yet according to Heath-Brown. Mathematica can found the following solution with some hint:

In[5]:= Reduce[a^3+b^3+(-159380)^3==39,{a,b},Integers]
Out[5]= (a==117367&&b==134476)||(a==134476&&b==117367)
POSTED BY: Shenghui Yang

There isn't, they checked a,b,c up to 10^14 in some clever way. (10 months CPU time (2007)).

a^3 +b^3 + c^3 == n

where

n \[Element] {33, 42, 74, 114, 165, 390, 579, 627, 633, 732, 795, 906, 921, 975}

are not solved yet! I would love to bruteforce it (using Mathematica) up to 10^15, to --perhaps-- eliminate some of them. Maybe you have some parallel computing power left over ;)

POSTED BY: Sander Huisman
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