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 Taking the Fifth (Posted on 2014-06-21)

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 No Rating

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 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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