Question 1208227
<pre>

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.

{{{x^2=x+1}}}

{{{x^2-x-1=0}}}

{{{x = (-(-1) +- sqrt( (-1)^2-4*(1)*(-1) ))/(2*(1)) }}}

{{{x = (1 +- sqrt(1+4))/2 }}}

{{{x = (1 +- sqrt(5))/2 }}} 

{{{x = 1/2 +- sqrt(5)/2 }}} 

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

{{{(a+b)^5}}}{{{""=""}}}{{{a^5+5a^4b^1+10a^3b^2+10a^2b^3+5ab^4+b^5}}}

{{{(1/2 + sqrt(5))^5 }}}{{{""=""}}}{{{(1/2)^5+(sqrt(5)/2)^5}}}{{{""=""}}}{{{(1/2)^5+5(1/2)^4(sqrt(5)/2)^1+10(1/2)^3y(sqrt(5)/2)^2+10(1/2)^2(sqrt(5)/2)^3+5(1/2)(sqrt(5)/2)^4+(sqrt(5)/2)^5}}}{{{""=""}}}
{{{1/32+5(1/16)(sqrt(5)/2)+10(1/8)(5/4)+10(1/4)(5sqrt(5)/8)+5(1/2)(25/16)+(25sqrt(5)/32)}}}{{{""=""}}}
{{{1/32+(5sqrt(5))/32+50/32+50sqrt(5)/32+125/32+(25sqrt(5))/32}}}{{{""=""}}}

{{{176/32+(80sqrt(5))/32}}}{{{""=""}}}{{{11/2+(5sqrt(5))/2}}}

also, {{{(1/2 - sqrt(5))^5 }}}{{{""=""}}}{{{11/2-(5sqrt(5))/2}}} 
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 x<sup>5</sup> + y<sup>5</sup>,
since x and y can be either one:

(1) {{{11/2+(5sqrt(5))/2}}}{{{""+""}}}{{{11/2+(5sqrt(5))/2}}}{{{""=""}}}{{{11+5sqrt(5)}}}

(2) {{{11/2+(5sqrt(5))/2}}}{{{""+""}}}{{{11/2-(5sqrt(5))/2}}}{{{""=""}}}{{{11}}}

(3) {{{11/2-(5sqrt(5))/2}}}{{{""+""}}}{{{11/2+(5sqrt(5))/2}}}{{{""=""}}}{{{11}}}

(4) {{{11/2-(5sqrt(5))/2}}}{{{""+""}}}{{{11/2-(5sqrt(5))/2}}}{{{""=""}}}{{{11-5sqrt(5)}}}

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

{{{11+5sqrt(5)}}}, {{{11}}}, and {{{11-5sqrt(5)}}}

Edwin</pre>