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Solution |
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![{(x+x\epsilon)}^n-x^n-{(x\epsilon)}^n=x^n [{(1+\epsilon)}^n-1-\epsilon^n] \\ =x^n {(-\epsilon^2)}^n -1 -\epsilon^n ....](http://s0.wp.com/latex.php?latex=%7B%28x%2Bx%5Cepsilon%29%7D%5En-x%5En-%7B%28x%5Cepsilon%29%7D%5En%3Dx%5En+%5B%7B%281%2B%5Cepsilon%29%7D%5En-1-%5Cepsilon%5En%5D+%5C%5C+%3Dx%5En+%7B%28-%5Cepsilon%5E2%29%7D%5En+-1+-%5Cepsilon%5En+....+&bg=ffffff&fg=333333&s=0 "{(x+x\epsilon)}^n-x^n-{(x\epsilon)}^n=x^n [{(1+\epsilon)}^n-1-\epsilon^n] \\ =x^n {(-\epsilon^2)}^n -1 -\epsilon^n ....") 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. |
is divisible by 
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is an odd number not divisible by 
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if 
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