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/sci/ - Science & Math

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10751723 No.10751723 [Reply] [Original]

3 cubed is 27 and 4 is 64. That makes the difference 37, which is prime. 6=216 7=343 difference is 127 which prime.

>> No.10751740

>>10751723
6^3 - 5^3 = 216 - 125 = 91 = 7 * 13. Sorry!

>> No.10751741

>>10751723
(n+1)^3 - n^3
(n+1)(n^2 + 2n + 1) - n^3
n^3 + 3n^2 + 3n + 1 - n^3
3n^2 + 3n + 1

Surely this doesn’t produce primes for all n

>> No.10751744

>>10751723
4^3-2^3

>> No.10751745

>>10751740
Is there a pattern for when it does happen to be prime? Has this been researched?

>> No.10751746

>>10751740
one also has 8^3 - 7^3 = 512 - 343 = 169 = 13*13.
plenty of 13s!
just in case it's not obvious, you're dealing with (n+1)^3 - n^3 = 3n^2 + 3n + 1, which of course will not be always prime valued on positive integers cause it's a parabola.

>> No.10751749
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10751749

>>10751744
OP obviously meant consecutive, moron.
>>10751745
feel free to look into the number theory behind when quadratics are prime valued but i doubt there's much special about 3n^2 + 3n + 1.

>> No.10751755

>>10751749
That parabola produced odd numbers no matter the input. That certainly helps

>> No.10751779

in case some basic facts aren't clear:
of course there will be no factors of 2 since n^2 + n is always even (if n is odd, then n^2 is odd, so their sum is even) so 3(n^2 + n) + 1 is always odd. it's also always 1 mod 3. so it's always 1 mod 6.
how about divisibility by 5? can 3(n^2 + n) be 4 mod 5? well, 3*0 = 0, 3*1 = 3, 3*2 = 6 = 1, 3*3 = 9 = 4, 3*4 = 12 = 2 mod 5. so we'd need that n^2 + n be 3 mod 5.
if n is 0 mod 5, so is n^2 + n.
if n is 1 mod 5, then n^2 + n is 2.
if n is 2 mod 5, then n^2 + n is 4 + 2 is 1.
if n is 3 mod 5, then n^2 + n is 9 + 3 is 2.
if n is 4 mod 5, then n^2 + n is 0.
so this n^2 + n can never be 3 mod 5. this shows that 3n^2 + 3n + 1 is never a multiple of 5.
thus the lowest prime factor of 3n^2 + 3n + 1 is always going to be 7, which is why it looks promising to begin with.

>> No.10751797

For the people wondering, this is a conjecture from a paper from like 50 years ago. Completely unsolved. It asks which polynomials produce infinitely many primes. The conjecture is irreducible polynomials with relatively prime coefficients and some other tiny restrictions.

Literally impossible to prove this beyond linear polynomials.

>> No.10751798

>>10751779
i'm curious about 11. is 3(n^2 + n) ever 10 mod 11? only if n^2 + n is 40 = 7 mod 11.
something something roots of n^2 + n - 7 in this field.
seems like it has no roots, so the thing isn't ever divisible by 11 either.
https://www.desmos.com/calculator/bcncyqetyt

>> No.10751817

>>10751745
It will never be prime if they are more than 1 apart.
x^3 - y^3 = (x-y)(x^2 +xy +y^2)

If they are 1 apart the problem reduces to studying the quadratic form x^2 + xy +y^2