Question 1208227
.
If x^2 = x+1 and y^2 = y+1, find x^5 + y^5.
~~~~~~~~~~~~~~~~~~~~



        I thought if there is a way to avoid long stupid calculations,

                        and finally got an idea.

        It seems a robust and elegant and deserves a separate presentation.



<pre>
From given equations, we have (after reducing to standard form quadratic equations)

    x^2 - x - 1 = 0,

    {{{x[1]}}} = {{{(1 + sqrt(5))/2}}},     (1)

    {{{x[2]}}} = {{{(1 - sqrt(5))/2}}},     (2)

    {{{y[1]}}} = {{{(1 + sqrt(5))/2}}},     (3)

    {{{y[2]}}} = {{{(1 - sqrt(5))/2}}}.     (4)


    +---------------------------------------------------------+
    |     The idea is that given equations allow to reduce    |
    |           the degrees of  x^5  and  y^5                 |
    |   step by step until we get linear binomial in x and y. |
    +---------------------------------------------------------+


I will write first step in all details. It is

    x^5 = x^2*x^3 = here I replace x^2 by (x+1) and continue = (x+1)*x^3 = x^4 + x^3.


So, we reduced the monomial x^5 to the polynomial  x^4 + x^3 of degree 4.

Next steps I will use standard equivalent transformations of polynomials,
will extract x^2 everywhere where possible and replace it by (x+1).


I will write my transformations in one long line. In order for do not make breaks
to explain every time that "here I replace x^2 by (x+1) and continue", I will use
special sign of doubled equality " = = ".


    +------------------------------------------------------+
    |   So, every time as you see this sign " = = ", read  |
    |   it as "here I replace x^2 by (x+1) and continue".  |
    +------------------------------------------------------+


Let's go.


    x^5 = x^2*x^3 = = (x+1)*x^3 = x^4 + x^3 = x^2*(x^2+x) = = (x+1)*(x^2+x) = x^3 + x^2 + x^2 + x =

        = x*3 + 2x^2 + x = x^2*x + 2x^2 + x = = (x+1)*x + 2*(x+1) + x = x^2 + x + 3x + 2 = = (x+1) + 4x + 2 =

        = 5x + 3.


Thus, we have  

    x^5 = 5x + 3.     (5)


Similarly,

    y^5 = 5y + 3.     (6)


So,  x^5 + y^5 = 5x + 5y + 6.


Now substitute values  {{{x[1]}}},  {{{x[2]}}},  {{{y[1]}}},  {{{y[2]}}}  from (1) - (4)  into this simple formula and get


    {{{x[1]^5}}} + {{{y[1]^5}}} = {{{5*((1+sqrt(5))/2)+5*((1+sqrt(5))/2)+6}}} = {{{5+5*sqrt(5)+6}}} = {{{11+5*sqrt(5)}}};

    {{{x[2]^5}}} + {{{y[1]^5}}} = {{{5*((1-sqrt(5))/2)+5*((1+sqrt(5))/2)+6}}} = 5 + 6 = 11;

    {{{x[1]^5}}} + {{{y[2]^5}}} = {{{5*((1+sqrt(5))/2)+5*((1-sqrt(5))/2)+6}}} = 5 + 6 = 11;

    {{{x[2]^5}}} + {{{y[2]^5}}} = {{{5*((1-sqrt(5))/2)+5*((1-sqrt(5))/2)+6}}} = {{{5-5*sqrt(5)+6}}} = {{{11-5*sqrt(5)}}}.
</pre>

At this point, the problem is solved completely.


The given equations allow to start (to turn on) the process of reducing degrees 
in the formulas, which quickly leads to the desired result.


It makes the solution straightforward, easy and elegant.



I hope that after this my solution you will see the problem from a completely 
different perspective and from a completely different angle of view.