How to Generate Cubes without Cubing: A Proof Method

[Charles H]

TL;DR

The difference between the cubes or any 2 consecutive integers is always a prime number.

I have tested the above proposition thru the integer 90, and have found that the proposition holds true. I'm not sure of the method to prove that this would always be true. Any help, criticism, or proof is welcome.

 

[fresh_42]

Hello and !

Do you mean the difference between $6^3-5^3=216-125=91=7\cdot 13?$

What you can prove is, that it is always an odd number. Do you know how to do this?

 

[FactChecker]

Sorry. There are a great many exceptions below 90. I assume that you tested it to 90 using a computer program of some sort. If you are interested, we might help you to find the bug in the program.

 

[PeroK]

TL;DR Summary: The difference between the cubes or any 2 consecutive integers is always a prime number.

Specifically:
$$(n+1)^3 - n^3 = 3n^2 + 3n + 1$$So the sequence is:
$$1, 7, 19,37,61, 91(*), 127, 169(*), 217(*), 271, 331, 397, 469(*) \dots $$Those with an (*) are not prime.

 

[Vanadium 50]

t is always an odd number

All odd numbers greater than one are prime. 3, 5, 7, 9 (experimental error), 11, 13...

(n+1)^3 - n^3 = 3h^2 + 3n + 1
(n+1)^3 - n^3 = 3n(n+1)+ 1

Now that does not factor, so it will be the case that it is sometimers prime, It's also, as pointed out, never divisible by 2 and by inspection never divisible by 3.

But suppose n = 7m + 1

(n+1)^3 - n^3 = 3(7m+1)(7m + 2)+ 1
(n+1)^3 - n^3 = 147m^2 + 63m + 7
(n+1)^3 - n^3 = 7(21m^2 +9m + 1)

which is by inspection always divisible by 7.

 

[PeroK]

PS if we let $d(n) = 3n^2 + 3n + 1$, then the difference in this sequence is:
$$d(n+1) - d(n) = 6(n+1)$$So, you can generate the sequence of consecutive integer cubes by starting at $1$ and adding the next integer multiple of $6$ each time.

 

[bob012345]

It appears this proposition is not ready for prime time...

 

[PeroK]

But suppose n = 7m + 1

(n+1)^3 - n^3 = 3(7m+1)(7m + 2)+ 1
(n+1)^3 - n^3 = 147m^2 + 63m + 7
(n+1)^3 - n^3 = 7(21m^2 +9m + 1)

which is by inspection always divisible by 7.

And, if $n = 19m + 2$, then $d(n)$ is divisible by $19$.

If $n = 37m + 3$, then $d(n)$ is divisible by $37$.

If $n = 61m + 4$, then $d(n)$ is divisible by $61$.

If $n = 91m + 5$, then $d(n)$ is divisible by $91$.

Etc.

 

[Charles H]

Hello and !

Do you mean the difference between $6^3-5^3=216-125=91=7\cdot 13?$

What you can prove is, that it is always an odd number. Do you know how to do this?

Thanks. That makes me feel really dumb. I overlooked the result on several examples. And, yes, I know how to prove it will always be odd.

 

[Vanadium 50]

Etc.

Of course you and I both know that there are not just these examples but an infinite number of them.

 

[bob012345]

PS if we let $d(n) = 3n^2 + 3n + 1$, then the difference in this sequence is:
$$d(n+1) - d(n) = 6(n+1)$$So, you can generate the sequence of consecutive integer cubes by starting at $1$ and adding the next integer multiple of $6$ each time.

I don't follow that statement. If we let $D(n)=d(n+1)-d(n)$ then $D(n+1)-D(n)=6$ but does it mean anything?

 

[PeroK]

I don't follow that statement. If we let $D(n)=d(n+1)-d(n)$ then $D(n+1)-D(n)=6$ but does it mean anything?

It gives you a quick way to generate the cubes without multiplication. The above also follows from:

If $f(x) = x^3$, then $f'''(x) = 6$.

 

[bob012345]

It gives you a quick way to generate the cubes without multiplication. The above also follows from:

If $f(x) = x^3$, then $f'''(x) = 6$.

I'm sorry, I just don't see how. Could you work out a few terms please?

 

[PeroK]

I'm sorry, I just don't see how. Could you work out a few terms please?

How to generate the cubes without cubing!

1
1 + 6 = 7; 1 + 7 = 8
7 + 12 = 19; 8 + 19 = 27
19+18 = 37; 27+37 = 64
37 + 24 = 61; 64 + 61 = 125
...