document.write( "Question 847697: Prove or disporove. If (n^3)+2 is odd then n^2 is odd where n is an element of all integers. \n" ); document.write( "

Algebra.Com's Answer #510640 by jim_thompson5910(35256) $\"\"$ $\"About$

You can put this solution on YOUR website!
This isn't a formal proof, but it's just the basic outline (to get you thinking in the right direction)\r
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\n" ); document.write( "\n" ); document.write( "If n^3 + 2 is odd, then n^3 must be odd. Why? Because the +2 makes you jump from even to even or from odd to odd. An example would be 7+2 = 9. Notice how we go from 7 to 9 when we increase by 2.\r
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\n" ); document.write( "\n" ); document.write( "So we know n^3 is odd. It turns out that if you cube any even number, you get an even number. Similarly, if you cube any odd number, you get an odd number. So because n^3 is odd, this must mean that n has to be odd. \r
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\n" ); document.write( "\n" ); document.write( "We know n is odd. So n = 2k + 1 for some integer k. Square this to get n^2 = 4k^2 + 4k + 1 ---> n^2 = 2(2k^2 + 2k) + 1 ----> n^2 = 2m + 1 for some integer m (m = 2k^2 + 2k). This shows us that if n is odd, then n^2 is odd.\r
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\n" ); document.write( "\n" ); document.write( "Let me know if this helps or not. \n" ); document.write( "

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