|1.Considering the given expression as a polynomial in , let us put . We see that at the polynomial vanishes (for any ). Therefore our polynomial is divisible by . Similarly, it is divisible by as well. Thus the polynomial is divisible by .
To prove that it is divisible by , put . It is evident that for odd n we have : for .
Consequently, our polynomial is divisible by . It only remains to prove the divisibility of the polynomial by , which also be written as where .
For this purpose it only remains to replace first by and then by to make sure that with these substitutions the polynomial vanishes. Since, by hypothesis, is not divisible by 3, it follows that , for every , in which is not acceptable since is odd from the problem. At the polynomial attains the following value
since () substituting we get
Likewise we prove that at the polynomial vanishes as well, and consequently, its by divisibility by is proved.
|2.To prove the second statement, let us proceed as follows. Let the quantities be the roots of a cubic equation . Then by virtue of the known relations between the roots of an equation and its coefficients we have or and .
Thus, are the roots of the equation where and
Put . Among successive values of , there exist the relationship ,: being equal to zero.
Let us prove that is divisible by if using the method of mathematical induction. Suppose is divisible by and prove that then is also divisible by .
So, using this relation we get that
Since, by supposition, is divisible by , it suffices to prove that is divisible by . Thus we only have to prove than is divisible by if , we easily prove our assertion. And so, assuming that is divisible by , we have proved that (from induction) is also divisible by . Consequently for any is divisible by .
Now it only remains to prove its divisibility by and by , which is quite elementary.