Prove by direct method that for any integer ‘n’, n3 – n is always even.
We have given, n3-n
Let us Assume, n is even
Let n=2k, where k is natural number
n3-n=(2k)3-(2k)
n3-n=2k(4k2-1)
let k(4k2-1)=m
n3-n=2m
Therefore, (n3-n) is even.
Now, Let us Assume n is odd
Let n=(2k+1), where k is natural number
n3-n=(2k+1)3-(2k+1)
n3-n= (2k+1)[(2k+1)2-1]
n3-n= (2k+1)[(4k2+4k+1-1)]
n3-n= (2k+1)[(4k2+4k)]
n3-n= 4k(2k+1)(k+1)
n2-n= 2.2k(2k+1)(k+1)
let λ=2k(2k+1)(k+1)
n3-n=2λ
therefore, n3-n is even.
Hence, n3-n is always even
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