document.write( "Question 1033752: Prove that there are no positive integers x, y, z such that
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Algebra.Com's Answer #648399 by ikleyn(52835) $\"\"$ $\"About$

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document.write( "Consider, one after one, these steps.\r\n" );
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document.write( "1.  Assume that both x and y are multiple of 3, and get a contradiction.  (It is easy)\r\n" );
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document.write( "       Hence, both x and y can not be multiple of 3 simultaneously.\r\n" );
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document.write( "2.  Assume that x is multiple of 3, and get a contradiction.  (It is easy)\r\n" );
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document.write( "3.  Assume that y is multiple of 3, and get a contradiction.  (It is easy)\r\n" );
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document.write( "       Hence, nor x neither y can be multiple of 3 even separately.\r\n" );
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document.write( "4.  Hence,  x  must have the remainder 1 or 2 when divided by 3: x = 3a + r,\r\n" );
document.write( "       where  r = 1 or 2.\r\n" );
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document.write( "5.  Same for y:  y  must have a remainder 1 or 2 when divided by 3:  y = 3b + s,\r\n" );
document.write( "       where  s = 1 or 2.\r\n" );
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document.write( "6.  Hence,      when squared,  has a reminder 1 when divided by 3.\r\n" );
document.write( "    Similarly,  when squared,  has a reminder 1 when divided by 3.\r\n" );
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document.write( "    It implies that  gives the remainder 2 when divided by 3.\r\n" );
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document.write( "    But the right side of the equation is a multiple of 3. \r\n" );
document.write( "    Contradiction.\r\n" );
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document.write( "So, there is no way for the given equation to be valid.\r\n" );
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