document.write( "Question 548928: Determine all positive integers n for which $\"2%5En+%2B+1\"$ is divisible by 3. \n" ); document.write( "

Algebra.Com's Answer #357335 by mathie123(224) $\"\"$ $\"About$

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We can test for the first few values of n (n=0, n=1, n=2, n=3, n=4...) and see a pattern emerging. It seems that when n is odd, $\"2%5En%2B1\"$ is divisible by 3. \r
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\n" ); document.write( "\n" ); document.write( "We can use induction to prove that this is true for all even values of n. \r
\n" ); document.write( "\n" ); document.write( "We have tried base cases, so it holds for the base cases. \r
\n" ); document.write( "\n" ); document.write( "Since any even number can be written as 2k where k is an integer, we will assume that for all values up to k, 2^{2*k}+1=3*m (m is an integer, meaning it is divisble by 3.
\n" ); document.write( "This can be rewritten as:
\n" ); document.write( "2^{2(k)}=3m-1\r
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\n" ); document.write( "\n" ); document.write( "We must now prove that 2^{2(k+1)}+1 is divisible by 3.
\n" ); document.write( "2^{2(k+1)}+1
\n" ); document.write( "=2^(2k)*2^2+1
\n" ); document.write( "=4*(3m-1)+1
\n" ); document.write( "=12m-4+1
\n" ); document.write( "=12m-3
\n" ); document.write( "Now since both 12 and -3 are divisible by 3, then this whole value is divisible by three.. so our proof is done \r
\n" ); document.write( "\n" ); document.write( "Hopefully this helps, let me know if you are still unsure :) \n" ); document.write( "

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