SOLUTION: Show that if n is odd integer, then n^3-n is a multiple of 24.

 =  =  = .


You have a product of three consecutive integers. 


One of them is a multiple of 3.


If n is odd, then both (n-1) and (n+1) are even, i.e. are multiples of 2.


Moreover, one of these two is a multiple of 4.


Therefore, the entire product is a multiple of 2*3*4 = 24.


QED.

She assumed that

"If n is odd, then both (n-1) and (n+1) are even, i.e. are multiples
of 2."

I agree with that without proof.  However when she says:

"Moreover, one of these two is a multiple of 4."

although that is true, I think it must be proved before she can use it
legitimately.

I do an induction proof:



Since n is odd, let n = 2m+1 where m is any integer




Then we have to prove that:

then 8m^3+12m^2+4m}}} is a multiple of 24 for any integer m

If m=1

4(1)(1+1)(2*1+1) = 4(2)(2+1) = 4(2)(3) = 24

and 24 is a multiple of 24.


Assume that m=k is such that  = 24 times some integer,

Now we consider   .  When we expand that
out we get 



The first parenthetical expression is what we assumed was a multiple of 24 
and the second parenthetical expression is a multiple if 24 and the sum
of two multiples of 24 is a multiple of 24.

Thus the theorem is proved by induction.

Edwin