document.write( "Question 1165198: Prove that n^5−n is divisible by 30 for n∈Z \n" ); document.write( "
Algebra.Com's Answer #789718 by Edwin McCravy(20060)\"\" \"About 
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document.write( "\"n%5E5+-+n\"\r\n" );
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document.write( "We completely factor:\r\n" );
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document.write( "\"n%28n%5E4-1%29\"\r\n" );
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document.write( "\"n%28n%5E2-1%29%28n%5E2%2B1%29\"\r\n" );
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document.write( "\"n%28n-1%29%28n%2B1%29%28n%5E2%2B1%29\"\r\n" );
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document.write( "To be divisible by 30, the above must be divisible by 2, 3, and 5\r\n" );
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document.write( "3 consecutive integers n-1, n, and n+1 are always such that one of\r\n" );
document.write( "them is divisible by 2 and another is divisible by 3. thus n5-n is always\r\n" );
document.write( "divisible by 6.\r\n" );
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document.write( "So we only need to show that one of the four factors is divisible by\r\n" );
document.write( "5.\r\n" );
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document.write( "For any integer n, every sequence of n consecutive integers can be written\r\n" );
document.write( "as nk+a,nk+a+1,nk+a+2,...,nk+a+n-1, for some integers k and a.\r\n" );
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document.write( "We can let n=5, a=-3\r\n" );
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document.write( "Then every sequence of 5 integers can be written 5k-3, 5k-2, 5k-1, 5k, 5k+1 \r\n" );
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document.write( "n is divisible by 5 if n=5k\r\n" );
document.write( "n-1 is divisible by 5 if n=5k+1 because (5k+1)-1 = 5k \r\n" );
document.write( "n+1 is divisible by 5 if n=5k-1 because (5k-1)+1 = 5k\r\n" );
document.write( "n²+1 is divisible by 5 if n=5k-2 because (5k-2)²+1 = 25k²+10k+4+1 =\r\n" );
document.write( "      25k²+10k+5 = 5(5k²+2k+1)\r\n" );
document.write( "n²+1 is divisible by 5 whenever n=5k-3 because (5k-3)²+1 = 25k²-30k+9+1 =\r\n" );
document.write( "      25k²+15k+10 = 5(5k²+3k+2)\r\n" );
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document.write( "So for every sequence of 5 consecutive integers, one of the 4 factors of\r\n" );
document.write( "n5-n is divisible by 2,3,5, which means n5-n is always\r\n" );
document.write( "divisible by 30.\r\n" );
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document.write( "[PROVED]\r\n" );
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document.write( "Edwin
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