Question 517660
<pre>
We are to prove

if x<sup>5</sup>=x<sup>4</sup>+1, then x³=x+1

or

if x<sup>5</sup>-x<sup>4</sup>-1=0, then x³-x-1=0

or we are to prove:

if f(x) = x<sup>5</sup> - x<sup>4</sup> - 1 = 0  and g(x) = x³ - x - 1 = 0

then all zeros of f(x) are zeros of g(x).  

This is likely not true for all complex zeros of f(x), since a 5th degree
polynomial has 5 zeros (counting multiplicities) whereas a 3rd degree
polynomial has only 3.  However it may be true for x real.


Let's consider the case when x is a real number.

By Descartes' rule of signs, both f(x) and g(x) have 1 positive zero.  

f(x) has no negative solutions. g(x) either has 2 or no negative zeros.
They could have the same positive zero if g(x) were a factor of f(x).
So we see if this is the case by long division of f(x)÷g(x):
                                                                            
                <u>                   x² -  x + 1</u>
x³ + 0x² - x - 1)x<sup>5</sup> -  x<sup>4</sup> + 0x³ + 0x² - 0x - 1
                 <u>x<sup>5</sup> + 0x<sup>4</sup> -  x³ -  x²</u>
                      -x<sup>4</sup> +  x³ +  x² - 0x
                      <u>-x<sup>4</sup> - 0x³ +  x² +  x</u>
                             x³ + 0x² -  x - 1
                             <u>x³ + 0x² -  x - 1</u>
                                             0
Yes indeed we get a 0 remainder, so

x<sup>5</sup>-x<sup>4</sup>-1 = (x³-x-1)(x²-x+1)

f(x) = g(x)·(x²-x+1)

x²-x+1 has only conjugate complex imaginary zeros {{{(1 +- i*sqrt(5))/2}}}, so if
x = the one and only real zero of f(x), the right side is also 0, and x²-x+1 is
not 0 since it has only imaginary zeros  {{{(1 +- i*sqrt(5))/2}}}, so x must also
be the real zero of g(x).  So the proposition is true for x real.

However it is not true for x complex imaginary, for the zeros
{{{(1 +- i*sqrt(5))/2}}} of x²-x+1 are zeros of f(x), but are not zeros of g(x).  

Edwin</pre>