document.write( "Question 517660: prove that if x^5=x^4+1, then x^3=x+1 \n" ); document.write( "
Algebra.Com's Answer #344880 by Edwin McCravy(20060)\"\" \"About 
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document.write( "We are to prove\r\n" );
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document.write( "if x5=x4+1, then x³=x+1\r\n" );
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document.write( "or\r\n" );
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document.write( "if x5-x4-1=0, then x³-x-1=0\r\n" );
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document.write( "or we are to prove:\r\n" );
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document.write( "if f(x) = x5 - x4 - 1 = 0  and g(x) = x³ - x - 1 = 0\r\n" );
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document.write( "then all zeros of f(x) are zeros of g(x).  \r\n" );
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document.write( "This is likely not true for all complex zeros of f(x), since a 5th degree\r\n" );
document.write( "polynomial has 5 zeros (counting multiplicities) whereas a 3rd degree\r\n" );
document.write( "polynomial has only 3.  However it may be true for x real.\r\n" );
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document.write( "Let's consider the case when x is a real number.\r\n" );
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document.write( "By Descartes' rule of signs, both f(x) and g(x) have 1 positive zero.  \r\n" );
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document.write( "f(x) has no negative solutions. g(x) either has 2 or no negative zeros.\r\n" );
document.write( "They could have the same positive zero if g(x) were a factor of f(x).\r\n" );
document.write( "So we see if this is the case by long division of f(x)÷g(x):\r\n" );
document.write( "                                                                            \r\n" );
document.write( "                                   x² -  x + 1\r\n" );
document.write( "x³ + 0x² - x - 1)x5 -  x4 + 0x³ + 0x² - 0x - 1\r\n" );
document.write( "                 x5 + 0x4 -  x³ -  x²\r\n" );
document.write( "                      -x4 +  x³ +  x² - 0x\r\n" );
document.write( "                      -x4 - 0x³ +  x² +  x\r\n" );
document.write( "                             x³ + 0x² -  x - 1\r\n" );
document.write( "                             x³ + 0x² -  x - 1\r\n" );
document.write( "                                             0\r\n" );
document.write( "Yes indeed we get a 0 remainder, so\r\n" );
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document.write( "x5-x4-1 = (x³-x-1)(x²-x+1)\r\n" );
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document.write( "f(x) = g(x)·(x²-x+1)\r\n" );
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document.write( "x²-x+1 has only conjugate complex imaginary zeros \"%281+%2B-+i%2Asqrt%285%29%29%2F2\", so if\r\n" );
document.write( "x = the one and only real zero of f(x), the right side is also 0, and x²-x+1 is\r\n" );
document.write( "not 0 since it has only imaginary zeros  \"%281+%2B-+i%2Asqrt%285%29%29%2F2\", so x must also\r\n" );
document.write( "be the real zero of g(x).  So the proposition is true for x real.\r\n" );
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document.write( "However it is not true for x complex imaginary, for the zeros\r\n" );
document.write( "\"%281+%2B-+i%2Asqrt%285%29%29%2F2\" of x²-x+1 are zeros of f(x), but are not zeros of g(x).  \r\n" );
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document.write( "Edwin
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