document.write( "Question 984151: I would appreciate your help!\r
\n" ); document.write( "\n" ); document.write( "If k stands for an integer, then is it possible for k^2+k to stand for an odd integer? Explain.
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Algebra.Com's Answer #604936 by reviewermath(1029)\"\" \"About 
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If k stands for an integer, then is it possible for k^2+k to stand for an odd integer? Explain.\r
\n" ); document.write( "\n" ); document.write( "The answer is \"NO\".
\n" ); document.write( "Case 1: k is an odd integer
\n" ); document.write( "For some integer p, k = 2p + 1
\n" ); document.write( "So, k + 1 = 2p + 2 = 2(p+1) = 2q for some integer q
\n" ); document.write( "k^2 + k = k(k+1) = k(2q) = 2kq = 2r for some integer r.[Integers are closed under multiplication]
\n" ); document.write( "Therefore, if k is an odd integer, then k^2 + k is an even integer\r
\n" ); document.write( "\n" ); document.write( "case 2: k is an even integer
\n" ); document.write( "For some integer p, k = 2p
\n" ); document.write( "So, k + 1 = 2p + 1
\n" ); document.write( "k^2 + k = k(k+1) = 2p(2p+1) , multiple of 2
\n" ); document.write( "Therefore, if k is an even integer, then k^2 + k is an even integer
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