Club409

i can reason it out for n=3 i'm sure the other cases are similar but it would take a little more effort and i'm working on the back of an envelope.

you start with [(x+1)^3 - x^3] - [(y+1)^3 - y^3]
if you expand out the (^3)'s you get
[3(x^2)+3x] - [3(y^2) + 3y]
if you factor that slightly you get
3[x(x+1) - y(y+1)]
x(x+1) and y(y+1) are both odd for any x and y and the difference between odd numbers is an even number
so the overall difference is a multiple of 3 and 2 => multiple of 6
not very elegant but it makes sense to me

i'm guessing the other cases are pretty similar. it actually looks like for n=5 they're all multiples of 30 (2 3 and 5) so i'm going to wager that for 7th powers they're all multiples of 42 (237) and so on. maybe later i'll play around with the binomial coefficients to see if i can show it more clearly

x(x+1) would be even not odd. x(x+1) + 1 would be odd (you dropped the +1).

This is part of a larger pattern. The "divisible by six" isn't even always true (try a few n = 2 examples). I'm pretty sure it's going to relate to some factorial relationship but I don't know for sure yet.

I started on a general proof this morning but got hung up studying anatomy. I'll probably have something up at some point.

my bad on the even/odd had it written down on the envelope right transcribed incorrectly. but a difference between even numbers is still even so the rest still holds. it's breaking down for large numbers because of numerical precision. in double precision you're going to start losing accuracy when you subtract nearly equal numbers that are approximately 10^16 (208065^3 is around 9x10^15). the rule isn't breaking down the calculator is. if you try it in quadruple precision i bet it still works.

i tried to give it a shot but once i ran out of room on one side of the paper i gave up (in hind sight i should've used a calculator sooner). basically what steven said but a little extra...

when you expand the whole thing you always lose the first and last term (aka the highest x and y power and the integer (1)). when you factor back what is left you can always get it in terms of n*x*(x+1)*w - n*y*(y+1)*w where w is potentially any odd number. for instance:

n=5 n*x*(x+1)*(x^2+x+1) - same thing for y; w = (x^2+x+1)
then w = (x(x+1)+1) = always odd like smoov said
n=7 n*x*(x+1)*(x^2+x+1)^2 - same thing for y;
odd squared = always odd
n=9 is a mess but you can do it.

so basically it boils down to the factors in Stevens proof when dealing with n=3 it always comes down to n and x(x+1). therefore 23n always seems to hold true.

the last thing I would've liked to have tried further but didn't. it seems like arif's z is always a factor of 23n and the number of "unique factors". what i mean by that is for n=5 there are 3 "factors" (x (x+1) and (x^2+x+1). the same 3 "factors" for n=7. and then for n=9 there are 5 factors: x (x+1) and 3 different (x^2 + __x+___). I had excel pick random numbers (only 10) to test this theory for 9 and it seemed to work where z was always divisible by 5 (no idea why). I wanted to test it out for n=11 but there were 6 "factors" so that is pointless.

someone needs to get Gokhan in on this.