Question 1070276
<pre>
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:

{{{n^3-n=n^2(n-1)=n(n-1)(n+1)}}}

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

{{{n^3-n=(2m+1)(2m)(2m+2) = (2m+1)(2m)2(m+1) = 4m(m+1)(2m+1) = 8m^3+12m^2+4m}}}


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 {{{8k^3+12k^2+4k}}} = 24 times some integer,

Now we consider   {{{8(k+1)^3+12(k+1)^2+4(k+1)}}}.  When we expand that
out we get 

{{{8k^3+36k^2+52k+24}}}{{{""=""}}}{{{(8k^3+12k^2+4k)+(24k^2+48k+24)}}}{{{""=""}}}{{{(8k^3+12k^2+4k)+24(k^2+2k+1)}}}

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</pre>