Question 903349
Prove: If n is an integer, then n^2 + n^3 is an even number.

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Case A: n is an even integer


So n = 2k for some integer k



n = 2k


n^2 = (2k)^2


n^2 = 4k^2


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n = 2k


n^3 = (2k)^3


n^3 = 8k^3


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n^2 + n^3 = 4k^2 + 8k^3


n^2 + n^3 = 2*(2k^2 + 4k^3)


that proves n^2 + n^3 is even since 2 is a factor.


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Case B: n is an odd integer



n = 2m+1 for some integer m



n = 2m+1


n^2 = (2m+1)^2


n^2 = 4m^2 + 4m + 1


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n = 2m+1


n^3 = (2m+1)^3


n^3 = 8m^3+12m^2+6m+1


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n^2 + n^3 = 4m^2 + 4m + 1 + 8m^3+12m^2+6m+1


n^2 + n^3 = 8m^3+16m^2+10m+2


n^2 + n^3 = 2*(4m^3+8m^2+5m+1)


that proves n^2 + n^3 is even since 2 is a factor.


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That wraps up the proof because there are only two possible cases if n is an integer.