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# prove that for every even n,  n(n2+20)  is divisible by 48

Vikas TU
14149 Points
5 years ago
Since nn is even, we can write n=2kn=2k for some integer kk.Hint:n(n2+20)=2k((2k)2+20)=2k(4k2+20)=8k(k2+5)n(n2+20)=2k((2k)2+20)=2k(4k2+20)=8k(k2+5)Hence, 88 is a factor.Note further that one of kk or k2+5k2+5 must be even, and hence divisible by 22. Why?So now we know that 8⋅2=168⋅2=16 is a factor.All that remains to be shown is that 33 is also a factor.
mycroft holmes
272 Points
5 years ago
Writing n = 2k, the given expression = 8k3+40k = 8(k3-k)+48 k.

It remains to prove that k3-k = (k-1) k (k+1) is divisible by 6. But this is true as it is the product of 3 consecutive integers and hence divisible by 3! = 6 and that concludes the proof.