document.write( "Question 1138835: Determine all natural numbers n such that for all positive divisors d of n,
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Algebra.Com's Answer #756762 by Edwin McCravy(20060) $\"\"$ $\"About$

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Determine all natural numbers n such that for all positive divisors d of n,
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document.write( "n cannot be even because then it would have divisor d=1 and d+1 would be 2,\r\n" );
document.write( "which could not be a divisor of n+1 for n+1 would be odd.  So we know that n\r\n" );
document.write( "must be odd.\r\n" );
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document.write( "Every odd natural number n has positive divisor d=1, and d+1=2 will always \r\n" );
document.write( "be a divisor of the next integer n+1, for n+1 will be even. \r\n" );
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document.write( "Every odd integer n also has positive divisor d=n (itself), and d+1\r\n" );
document.write( "(which equals n+1) will always be a divisor of n+1 (itself).\r\n" );
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document.write( "We ask: Can n have divisors other that 1 and n itself?\r\n" );
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document.write( "Let's find out by dividing n+1 by d+1 by long division, assuming d is not 1.\r\n" );
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document.write( "      n/d_______\r\n" );
document.write( "d + 1) n  +   1\r\n" );
document.write( "       n  +  n/d\r\n" );
document.write( "            1-n/d = remainder = 0\r\n" );
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document.write( "The remainder here must be 0, so\r\n" );
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document.write( "So the answer is no, because d=n tells us that every divisor d other than 1 of n\r\n" );
document.write( "must be equal to then natural number n itself.  So n cannot have any divisors\r\n" );
document.write( "other than itself and 1.  The only natural numbers n that only have divisors\r\n" );
document.write( "1 and themselves are 1 and the odd primes.\r\n" );
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document.write( "Answer: n is either 1 or an odd prime number.\r\n" );
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document.write( "Edwin

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