Prove that r(r^2-1)(3r+2) is divisible by 24

[Aryamaan Thakur]

Can anyone help me with this divisibility problem.

My approach:-
24 = 2*2*2*3
Now,
This can be written as
(r-1)(r)(r+1)(3r+2)
There will be a multiple of 2 and a multiple of 3. But how to prove that there are more multiples of 2.

PLEASE REPLY FAST!

 

[fresh_42]

How about looking at the cases $r$ odd and $r$ even?

 

[mfb]

I would look at four cases of r.

 

[scottdave]

So for the arbitrary even case, you can let r = 2k, and substitute that in. What do you get there? Then for the arbitrary odd case, let r = (2k-1) and substitute that in. As you have already figured, there will always be a factor of 3 in there, so you only need to look for 8.

 

[Kaura]

I think a way to solve this might be the write out the sequences for

r
r^2-1
3r-2

and then separate for even and odd so that for example r^2-1 would give
Even - 3, 15, 35, 63
Odd - 0, 8, 24, 48

I played around a little with this method and started to see a pattern in the factors of the numbers at a given index of the sequence
If you can prove that and set of numbers from the same index of all three of the even or odd sequences will always include the prime factors of 24 then that should suffice

 

[Mark44]

Please, no more hints until the OP comes back. For someone looking for replies REAL FAST, it's odd that he hasn't responded for ten days.

 

[Kaura]

Perhaps he is traveling at 299,792,457 m/s