document.write( "Question 819841: Find couples m and n if mn-1|n^3-1. Numbers m and n are positive and elements of set N. \n" ); document.write( "
Algebra.Com's Answer #493656 by Edwin McCravy(20062)\"\" \"About 
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document.write( "(mn-1)|(n³-1)\r\n" );
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document.write( "Of course we must have n≧2 in all cases\r\n" );
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document.write( "If m=1, we have\r\n" );
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document.write( "(n-1)|(n³-1)\r\n" );
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document.write( "which of course is true, since n³-1 = (n-1)(n²+n+1)\r\n" );
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document.write( "If m=n², we have\r\n" );
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document.write( "(n³-1)|(n³-1)\r\n" );
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document.write( "which of course is true, since any natural number divides itself.\r\n" );
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document.write( "Now if n is a perfect square, say k², then we have\r\n" );
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document.write( "(mk²-1)|[(k²)³-1]\r\n" );
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document.write( "(mk²-1)|(k6-1)\r\n" );
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document.write( "Then we can have m=k, for then we'd have\r\n" );
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document.write( "(k³-1)|(k3-1)(k3+1)\r\n" );
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document.write( "So apparently there are 3 cases of solutions for n≧2\r\n" );
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document.write( "m=1, m=n², and the third case is when n is a perfect square, k², \r\n" );
document.write( "and m is its square root, k.\r\n" );
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document.write( "Edwin
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