Proving n² + n is Even

Real Analysis · Axiom Academy

EXAMPLE Proving n^2 + n is Even A direct proof by factoring: every integer's square plus itself is even. Prove that n^2 + n is even for every integer n . We give a direct proof : factor the expression, then argue from the structure of the factors. Why a Factor of 2 Is Guaranteed Integers alternate even, odd, even, odd… so among any two neighbors n and n+1 , exactly one is even (highlighted). That even neighbor hands the product n(n+1) its factor of 2 . Read it off: pick any two side-by-side cells. One is always blue (even). That holds no matter where you start on the line — which is exactly why n(n+1) is even for every integer n . Nice work — you proved n^2 + n is even for every integer n , using nothing but a clean factor and a parity fact. Factor first: n^2 + n = n(n+1) turns a sum into a product, where structure is easier to see. Consecutive integers: n and n+1 are neighbors, so exactly one of them is even. Even times anything is even: writing the even factor as 2m gives , a multiple of 2 . Result: n^2 + n is even for all integers n — positive, negative, or zero. A second route reaches the same conclusion by cases : if n = 2k then n^2+n = 2(2k^2+k) ; if n = 2k+1 then n^2+n = 2(2k^2+3k+1) — even either way. Factoring is shorter; case analysis is more systematic. Both are tools you will reuse all through this course.

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