Algebra.Com's Answer #846365 by ikleyn(52788)![\"\"](\"/images/stars/stars-13.gif\")  
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document.write( "Quadrilateral PQRS is cyclic and side PS = u is a
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document.write( "diameter of the circle. If PQ = QR = v, RS = w, and u, v,
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document.write( "and w are integers such that v does not equal w, prove that u cannot be a
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document.write( "prime number.
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document.write( " I will give another solution, different from Edwin, and more compact.\r
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document.write( "Make a sketch.\r\n" );
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document.write( "So, you have quadrilateral PQRS inscribed in the circle such that PS \r\n" );
document.write( "is a diameter of this circle; PQ = QR = v; RS = w; PS = u.\r\n" );
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document.write( "Draw the diagonal PR. Let PR = d (the length).\r\n" );
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document.write( "We have isosceles triangle PQR. Using the cosine law, we can write\r\n" );
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document.write( "  =  \r\n" );
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document.write( "or\r\n" );
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document.write( " =  . (1)\r\n" );
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document.write( "From the other hand, triangle PRS is the right triangle, since angle PRS \r\n" );
document.write( "leans on the diameter PS. Therefore,\r\n" );
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document.write( " =  . (2)\r\n" );
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document.write( "From (1) and (2), we can exclude and write\r\n" );
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document.write( " = . (3)\r\n" );
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document.write( "Since quadrilateral PQRS is inscribed in the circle, its opposite angles S and Q \r\n" );
document.write( "are supplementary: S + Q =  . Therefore, cos(Q) = -cos(S).\r\n" );
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document.write( "Hence, from (3) we have\r\n" );
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document.write( "  = . (4)\r\n" );
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document.write( "Next, cos(S) =  (from the right triangle PRS).\r\n" );
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document.write( "Thus\r\n" );
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document.write( "  = . (5)\r\n" );
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document.write( "Multiply both sides of (5) by \"u\". You will get\r\n" );
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document.write( "  =  .\r\n" );
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document.write( "Combine the terms with \"u\" on the right side; keep the remaining term  on the left side\r\n" );
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document.write( "  =  . (6)\r\n" );
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document.write( "In (6), right side is a multiple of \"u\". So, if \"u\" is a prime number, \r\n" );
document.write( "then left side is a multiple of \"u\".\r\n" );
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document.write( "Hence, EITHER \"u\" divides 2, OR \"u\" divides  , OR \"u\" divides \"w\".\r\n" );
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document.write( "It leads to contradiction, since\r\n" );
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document.write( " - if \"u\" divides 2, then u= 2; but then from the right triangle PRS the leg w\r\n" );
document.write( " must be shorter than the hypotenuse u, i.e. w= 1; and side v must be shorter \r\n" );
document.write( " than the diameter u= 2; so, in this case w = v, which is excluded in the problem.\r\n" );
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document.write( " - if \"u\" divides , then \"u\" divides \"v\"; but it is impossible, since \r\n" );
document.write( " \"v\" is less than \"u\" (the diameter).\r\n" );
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document.write( " - if \"u\" divides \"w\", it is also impossible, since \"w\" is less than \"u\" (the diameter).\r\n" );
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document.write( "These contradictions prove that \"u\" can not be a prime number.\r\n" );
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document.write( "Solved.\r
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