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Taking the Fifth (Posted on 2014-06-21) Difficulty: 3 of 5

For the most part, if z^5 is the sum of two positive cubes, then z^2 is itself the sum of positive cubes:
(z^2 = x^3 + y^3, so z^2*z^3 = z^5 = (z*x)^3 + (z*y)^3).

Hence, such solutions to z^5=x^3+y^3 are said to be 'trivial'

In this sense, 3549^5 is the non-trivial sum of two positive cubes. It is nevertheless possible to calculate the cubes without resort to brute force.

How, and what are they?

No Solution Yet Submitted by broll    
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Solution Solution | Comment 3 of 5 |
I started with the general equation N = a^3 + b^3.

Then N = (a+b)*(a^2-a*b+b^2).  Note that this implies a+b is a factor of N.  Also a+b >= cbrt(N).

The trick I saw was to let x=a+b and y=a-b.  Then 4*N = x*(x^2+3y^2).  Note that this implies x < cbrt(4*N).  (Since a+b=x then x is a factor of N.)

The equation can be rearranged into (4N-x^3)/(3x) = y^2.  The y^2 is important because any prime factors left after the division must occur an even number of times.

N=3549^5 and 3549=3*7*13^2.  Then x has at most three prime factors: 3, 7, and 13.  

The 3 in the denominator of (4N-x^3)/(3x) implies 3 must be a factor of x, and since x is a factor of N it can have at most five factors of 3.  But for y^2 to be an integer, only 3^2 and 3^4 are possible options.

7 does not necessarily need to be a factor of x. If 7 is a factor then it must occur 1, 3, or 5 times in order for y^2 to be an integer.  Similarily, 13 can occur 0,1,2,3,4,6,8, or 10 times.

This leaves a total of 64 possible values for x.  Take one value from each set {3^2,3^4}, {7^0, 7^1, 7^3, 7^5}, {13^0, 13^1, 13^2, 13^3, 13^4, 13^6, 13^8, 13^10} and calculate the product for a list of potential x.

The list can be refined further.  Recall the inequalities mentioned earlier, they create a range: cbrt(N) < x < cbrt(4*N).  This is actually a rather tight range since we are creating candidates by multiplying and the upper limit is only cbrt(4)=1.5874 times the lower limit.  

With our value of N the inequality becomes 825740 < x < 1310781. Of the 64 values exactly one fits in this range: x = 3^4*7^1*13^3 = 1245699.

Then y^2 = (4*3^5*7^5*13^10 - 3^12*7^3*13^9)/(3^5*7^1*13^3)
 = 4*7^4*13^7 - 3^7*7^2*13^6
 = 7^2*13^6*(4*7^2*13-3^7)
 = 7^2*13^6*361
Which implies y = 7*13^3*19 = 292201.  Then a and b are easily calculated to be a=768950 and b=476749.

  Posted by Brian Smith on 2017-05-01 23:55:28
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