20200213, 17:53  #1 
Mar 2018
17·31 Posts 
559
559=6^3+7^3 is the sum of two consecutive cubes
Is 559 (after 344) the smallest number k such that k is the sum of two positive cubes and k congruent to 0 mod 43? Are there any other k congruent to 0 mod 43 and sum of two positive cubes? Last fiddled with by enzocreti on 20200213 at 18:12 
20200213, 18:15  #2 
6809 > 6502
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Aug 2003
101×103 Posts
2^{3}·1,171 Posts 
Who cares?
Why might this not be significant? 
20200213, 21:55  #3 
Jan 2017
2^{2}·7 Posts 
x^3+y^3=(x+y)(x^2xy+y^2), so yes there are infinitely many numbers x and y such that x+y is divisible by 43, and thus x^3+y^3 as well by association.

20200213, 21:59  #4 
Mar 2018
17×31 Posts 
...
7^3+6^3=559
7^36^3=127 a Mersenne prime. Are there other Mersenne primes which are the difference of two positive cubes? 
20200213, 22:18  #5 
Jan 2017
2^{2}·7 Posts 
x^3y^3=(xy)(x^2+xy+y^2), so if xy=1, then y=x1 and the quadratic becomes x^2+x(x+1)+(x1)^2=x^2+x^2x+x^22x+1=3x^23x+1, so if 3x^23x+1 is a Mersenne prime, then x^3y^3 is also a Mersenne prime. Mersenne primes are always of the form 3n+1, so:
2^n1=3x^23x+1 2^n2=3x^23x (2^n2)/3=x^2x x^2x(2^n2)/3=0 So, using the quadratic formula, if (1)^24(1)((2^n2)/3)=1+4(2^n2)/3 with n a Mersenne prime exponent is a perfect square, then there is another Mersenne prime that is the difference of 2 cubes. This may seriously be one thing Enzo said that could turn into a legitimately difficult mathematical conjecture: Are there any more perfect squares of the form 1+4(2^n2)/3? Also, trivially, 7. Last fiddled with by NHoodMath on 20200213 at 22:24 
20200214, 05:47  #6  
Romulan Interpreter
Jun 2011
Thailand
5^{2}×7×53 Posts 
Quote:
And don't forget to keep us informed of the progress. Last fiddled with by LaurV on 20200214 at 05:48 

20200214, 05:58  #7  
Romulan Interpreter
Jun 2011
Thailand
243B_{16} Posts 
Quote:
Quote:
(In context, OEISA181123) Quote:

