Question 903349: Prove: If n is an integer, then n^2 + n^3 is an even number.
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! 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.
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