There may be another way to work this using the sums of the powers and 
powers of the sums of x and y, here it is using the binomial expansion.









 

 

y will be the same two values. Using the binomial theorem:









also,  
because the expansion will be like the above except that all the terms with
square roots will have - signs:

So there are potentially 4 answers for x5 + y5,
since x and y can be either one:

(1) 

(2) 

(3) 

(4) 

Since (2) and (3) are the same, there are only three answers:

, , and 

Edwin

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

    x^2 - x - 1 = 0,

     = ,     (1)

     = ,     (2)

     = ,     (3)

     = .     (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  ,  ,  ,    from (1) - (4)  into this simple formula and get


     +  =  =  = ;

     +  =  = 5 + 6 = 11;

     +  =  = 5 + 6 = 11;

     +  =  =  = .

An easier and more elegant way is to use



as a recurrence formula for higher powers of x








So  and 

Since  and   both have solutions 

 




There are 4 possibilities for x+y

 
 
 
 

Substituting in:







Moral: If you work at a problem long enough you will find the simplest
solution.  It's much easier when you have someone else's longer solution to
simplify.  And nobody is STUPID for failure to use the shortest possible
solution first.

Edwin