559

enzocreti · 2020-02-13, 17:53

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?

Uncwilly · 2020-02-13, 18:15

Who cares?

Why might this not be significant?

NHoodMath · 2020-02-13, 21:55

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.

enzocreti · 2020-02-13, 21:59

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?

NHoodMath · 2020-02-13, 22:18

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.

LaurV · 2020-02-14, 05:47

Quote:

Originally Posted by enzocreti

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

There are ONLY about 50 known mersenne primes. Be our guest to test all of them. It wouldn't be so difficult...

And don't forget to keep us informed of the progress.

LaurV · 2020-02-14, 05:58

Quote:

Originally Posted by NHoodMath

This may seriously be one thing Enzo said that could turn into a legitimately difficult mathematical conjecture:

Quote:

Are there any more perfect squares of the form 1+4(2^n-2)/3?

Don't feed this troll...

(In context, OEISA181123)

Quote:

Also, trivially, 7.

Huh?

2020-02-13, 17:53	#1
enzocreti Mar 2018 2×5×53 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 2020-02-13 at 18:12

2020-02-13, 21:59	#4
enzocreti Mar 2018 2·5·53 Posts	... 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?

2020-02-13, 22:18	#5
NHoodMath Jan 2017 3×11 Posts	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. Last fiddled with by NHoodMath on 2020-02-13 at 22:24

2020-02-13, 18:15	#2
Uncwilly 6809 > 6502 """"""""""""""""""" Aug 2003 101×103 Posts 9,787 Posts	Who cares? Why might this not be significant?

2020-02-13, 21:55	#3
NHoodMath Jan 2017 3·11 Posts	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.