document.write( "Question 349548: Prove the following statement.\r
\n" ); document.write( "\n" ); document.write( "If n=m^3-m for some integer m, then n is a multiple of 6.\r
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\n" ); document.write( "\n" ); document.write( "My work so far:\r
\n" ); document.write( "\n" ); document.write( "Suppose n=m^3 - m. Thus n=m(m-1)(m+1). First if m=2k for some integer k. Thus n=(2k)(2k-1)(2k+1)= 2 (k(2k-1)(2k+1)). Also if m=2k+1 for some integer k, then n=(2k+1)(2k)(2k+2)=2(k(2k+1)(2k+2))\r
\n" ); document.write( "\n" ); document.write( "Thus n is divisible by 2. \r
\n" ); document.write( "\n" ); document.write( "I would like to find out how n is also divisible by 3 because if n is divisible by 3 as well then n would be divisible by 6. \n" ); document.write( "
Algebra.Com's Answer #249859 by jsmallt9(3758)\"\" \"About 
You can put this solution on YOUR website!
If n=m^3-m for some integer m, then n is a multiple of 6.\r
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\n" ); document.write( "\n" ); document.write( "My work so far:\r
\n" ); document.write( "\n" ); document.write( "Suppose n=m^3 - m. Thus n=m(m-1)(m+1).
\n" ); document.write( "This is correct. From this point I would say
  • If m is an integer, then (m-1), m, and (m+1) are 3 consecutive integers
  • In any set of 3 consecutive integers, there will be a multiple of 2 and a multiple of 3.
  • Any number that is the product of a multiple of 2 and a multiple of 3 will be a multiple of 6.

\n" ); document.write( "This may not be particularly formal as a proof but the logic is perfectly sound.
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