document.write( "Question 1172305: Hi guys, this is a proofing type question.\r
\n" ); document.write( "\n" ); document.write( "There are three parts - I don't know how to answer the last one. Part 1 and 2 should help with the final part. They are summarised below.\r
\n" ); document.write( "\n" ); document.write( "Every odd number is one more or one less than a multiple of 4 \r
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\n" ); document.write( "\n" ); document.write( "The product of any two positive integers of the form \"4n%2B1\", n is positive integer, is also of the form \"4n%2B1\".\r
\n" ); document.write( "\n" ); document.write( "The actual question is:\r
\n" ); document.write( "\n" ); document.write( "Hence, prove by contradiction that any composite number of the form \"4n-1\" must have at least one prime factor of the form \"4n-1\".
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Algebra.Com's Answer #797327 by ikleyn(52803)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "So, the statement, which you want to prove, is THIS\r\n" );
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document.write( "   |  prove by contradiction that any composite number of the form (4n-1)   |\r\n" );
document.write( "   |  must have at least one prime factor of the form (4n-1).               |\r\n" );
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document.write( "Let N be a composite number of the form (4n-1).\r\n" );
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document.write( "Then it is a product of the odd prime numbers; the prime number 2 is not its divisor.\r\n" );
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document.write( "Let assume that all its prime divisors are of the form (4n+1).\r\n" );
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document.write( "Notice that the product of any two odd numbers of the form (4n+1) is the number of the form (4n+1).\r\n" );
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document.write( "    It can be proven by direct multiplication of the numbers of this form.\r\n" );
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document.write( "It implies that a product of ANY number of the primes of the form (4n+1) has the form (4n+1).\r\n" );
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document.write( "But our number N has the form (4n-1).\r\n" );
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document.write( "So, we got a CONTRADICTION, which proves the statement.\r\n" );
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