document.write( "Question 476630: \"if (2^k)-1 is a prime number then, prove that (2^(k-1))(2^k-1)is a perfect number.\"
\n" ); document.write( "(i know it is the definition but i want proof)
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Algebra.Com's Answer #326822 by MathLover1(20849)\"\" \"About 
You can put this solution on YOUR website!
Theorem:
\n" ); document.write( " If \"2k-1\" is a \"prime\" number, then \"2k-1%282k-1%29\" is a \"perfect\" number and every even perfect number has this form.\r
\n" ); document.write( "\n" ); document.write( "Proof: \r
\n" ); document.write( "\n" ); document.write( "Suppose first that \"p+=+2k-1\" is a \"prime\" number, and set \"n+=+2k-1%282k-1%29\". \r
\n" ); document.write( "\n" ); document.write( "To show \"n\" is \"perfect\" we need only show \"sigma%28n%29+=+2n\". Since \"sigma\" is multiplicative and \"sigma%28p%29+=+p%2B1+=+2k\", we know\r
\n" ); document.write( "\n" ); document.write( " \"sigma%28n%29+=+sigma%282k-1%29.sigma%28p%29+=++%282k-1%292k+=+2n\".\r
\n" ); document.write( "\n" ); document.write( "This shows that \"n\" is a \"perfect\" number.\r
\n" ); document.write( "\n" ); document.write( " On the other hand, \"suppose\"\"+n\" is any \"even\"\"+perfect\" number and write \"n\" as \"2k-1m\" where \"m\" is an \"odd\"\"+integer\" and \"k%3E2\". \r
\n" ); document.write( "\n" ); document.write( "Again sigma is multiplicative so\r
\n" ); document.write( "\n" ); document.write( " \"sigma%282k-1m%29+=+sigma%282k-1%29.sigma%28m%29+=+%282k-1%29.sigma%28m%29\".\r
\n" ); document.write( "\n" ); document.write( "Since \"n\" is \"perfect\" we also know that\r
\n" ); document.write( "\n" ); document.write( " \"sigma%28n%29+=+2n+=+2km\".\r
\n" ); document.write( "\n" ); document.write( "Together these two criteria give\r
\n" ); document.write( "\n" ); document.write( " \"2km+=+%282k-1%29.sigma%28m%29\",\r
\n" ); document.write( "\n" ); document.write( "so, \"2k-1\" divides \"2km\" hence \"2k-1\" divides \"m\", say \r
\n" ); document.write( "\n" ); document.write( "\"m+=+%282k-1%29M\". \r
\n" ); document.write( "\n" ); document.write( "Now substitute this back into the equation above and \"divide\" by \"2k-1\" to get \"2kM+=+sigma%28m%29\". \r
\n" ); document.write( "\n" ); document.write( "Since \"m\" and \"M\" are both \"divisors\" of \"m\" we know that\r
\n" ); document.write( "\n" ); document.write( " \"2kM+=+sigma%28m%29+%3E+m+%2B+M+=+2kM\",\r
\n" ); document.write( "\n" ); document.write( "so \"sigma%28m%29+=+m+%2B+M\". \r
\n" ); document.write( "\n" ); document.write( "This means that \"m\" is \"prime\" and its \"only\" two \"divisors\" are itself (\"m\") and one (\"M\"). \r
\n" ); document.write( "\n" ); document.write( "Thus \"m+=+2k-1\" is a \"prime\" and we have \"prove\" that the number \"n\" has the prescribed form. \r
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