document.write( "Question 1207339: Show that x^2 + x + 1 is prime. \n" ); document.write( "
Algebra.Com's Answer #845167 by ikleyn(52805)![]() ![]() You can put this solution on YOUR website! . \n" ); document.write( "Show that x^2 + x + 1 is prime. \n" ); document.write( "~~~~~~~~~~~~~~~~~~~~~~~~\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \r\n" ); document.write( "Consider the discriminant of this quadratic polynomial.\r\n" ); document.write( "\r\n" ); document.write( "The discriminant is\r\n" ); document.write( "\r\n" ); document.write( " d = b^2 - 4ac = 1^2 - 4*1*1 = -3.\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "The discriminant is negative, which means that the polynomial does not have real roots.\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "If this polynomial be a composite over real numbers, it would be a product of two linear \r\n" ); document.write( "binomials (ax+b)*(cx+d).\r\n" ); document.write( "\r\n" ); document.write( "But then it would have two real roots, -b/a and -c/d.\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "Thus we get a CONTRADICTION with the fact proven above that this polynomial has no real roots.\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "The contradiction PROVES that the polynomial is not a composite over real numbers.\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "Hence, it is a PRIME over real numbers.\r\n" ); document.write( "\r \n" ); document.write( "\n" ); document.write( "Solved.\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |