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?

Who cares?

Why might this not be significant?

x^3+y^3=(x+y)(x^2-xy+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.

...
 

7^3+6^3=559

7^3-6^3=127 a Mersenne prime.

Are there other Mersenne primes which are the difference of two positive cubes?

x^3-y^3=(x-y)(x^2+xy+y^2), so if x-y=1, then y=x-1 and the quadratic becomes x^2+x(x+-1)+(x-1)^2=x^2+x^2-x+x^2-2x+1=3x^2-3x+1, so if 3x^2-3x+1 is a Mersenne prime, then x^3-y^3 is also a Mersenne prime. Mersenne primes are always of the form 3n+1, so:
2^n-1=3x^2-3x+1
2^n-2=3x^2-3x
(2^n-2)/3=x^2-x
x^2-x-(2^n-2)/3=0
So, using the quadratic formula, if (-1)^2-4(1)(-(2^n-2)/3)=1+4(2^n-2)/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^n-2)/3?
Also, trivially, 7.

[QUOTE=enzocreti;537529]Are there other Mersenne primes which are the difference of two positive cubes?[/QUOTE]
There are ONLY about 50 known mersenne primes. Be our guest to test all of them. It wouldn't be so difficult... :whistle:
And don't forget to keep us informed of the progress.

[QUOTE=NHoodMath;537531]This may seriously be one thing Enzo said that could turn into a legitimately difficult mathematical conjecture:
[/QUOTE]
:shit-just-got-real:
[QUOTE]
Are there any more perfect squares of the form 1+4(2^n-2)/3?
[/QUOTE]
Don't feed this troll...
:dnftt:

(In context, OEISA181123)
[QUOTE]
Also, trivially, 7.[/QUOTE]
Huh?
:orly emu: