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Positive Perfect Cube Sum (Posted on 2015-08-15) Difficulty: 3 of 5
Determine the smallest value of k ≥ 2 such that 20172017 is the sum of precisely k positive perfect cubes and prove that no smaller value of k is in conformity with the conditions of the problem.

No Solution Yet Submitted by K Sengupta    
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solution | Comment 2 of 3 |
Assume there's a solution for k=2 so that a^3 + b^3 = 2017^2017.

LHS factors as (a+b)(a^2 - ab + b^2) and since 2017 is prime, each of the terms in parentheses will be divisible by 2017.

Squaring the first term and subtracting the second term gives 3ab which will also be divisible by 2017 and therefore each of a,b is divisible by 2017 as well.

Then a^3 + b^3 will have a factor of 2017 raised to a power which = 0 mod3.  But 2017 = 1 mod3 and dividing by the LH factor leaves that status unchanged.  Repeating this process gives finally p^3 + q^3 = 2017 but 2017 isn't the sum of two cubes as you can verify.

So k>2 and a little fiddling yields a solution for k=3 which is minimum.

a=11x, b=c=7x with x=2017^672 gives a^3 + b^3 + c^3 = x^3 * (1331 + 343 + 343) = (2017^2016) * 2017 = 2017^2017.

  Posted by xdog on 2015-08-18 23:46:46
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