document.write( "Question 1179358: Use mathematical induction to prove each statement is true for all positive integers n:
\n" ); document.write( "5^(n)-1 is divisible by 4
\n" ); document.write( "n^(2)-n is divisible by 2
\n" ); document.write( "

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\n" ); document.write( "Use mathematical induction to prove each statement is true for all positive integers n:
\n" ); document.write( "5^(n)-1 is divisible by 4
\n" ); document.write( "n^(2)-n is divisible by 2
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\n" ); document.write( "\n" ); document.write( "n|d means \"n divides d\"\r
\n" ); document.write( "\n" ); document.write( "n=0: \"+5%5E%28n%29-1+=+5%5E0-1+=+0\", 4 | 0
\n" ); document.write( "n=1: \"5%5E%28n%29-1+=+5%5E1-1+=+4\", 4 | 4\r
\n" ); document.write( "\n" ); document.write( "Assume true for n=k: i.e. \"4\" | \"%285%5Ek+-+1%29\" (hypothesis)\r
\n" ); document.write( "\n" ); document.write( "Let n=k+1:
\n" ); document.write( "\"5%5E%28k%2B1%29-1\"
\n" ); document.write( "= \"5%5E%28k%2B1%29-5%2B4\"
\n" ); document.write( "= \"5%2A5%5Ek-5%2B4\"
\n" ); document.write( "= \"5%285%5Ek-1%29%2B4\"\r
\n" ); document.write( "\n" ); document.write( "4 | \"%285%5Ek-1%29\" (by the hypothesis) so \"4+\" | \"+5%2A%285%5Ek-1%29\" as well.\r
\n" ); document.write( "\n" ); document.write( "Therefore, \"+4+\" | \"5%285%5Ek-1%29%2B4\" (if 4|P then 4|(P+4))
\n" ); document.write( "and the proof is complete.\r
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\n" ); document.write( "\n" ); document.write( "The other proof does not require induction \"n%5E2-n+\" is always even, which is divisible by 2. Follow the steps I did in first problem if you must have a proof by induction (show true for base case, assume hypothesis (n=k), and then show it leads to truth of the step case where n=k+1)\r
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