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Deductive proof that the sum of the squares of two consecutive even integers is a multiple of four but not of eight

Deductive proof that the sum of the squares of two consecutive even integers is a multiple of four but not of eight

5 marks
  1. Show that \((2n)^2+(2n+2)^2=4\left(2n^2+2n+1\right)\), where \(n\in\mathbb{Z}\).[2]
  2. Hence, or otherwise, prove that when any two consecutive even integers are squared and the results are added, the total is a multiple of \(4\) but is never a multiple of \(8\).[3]

Solution:

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Answer:(a) \((2n)^2+(2n+2)^2=4\left(2n^2+2n+1\right)\) (b) the sum is \(4\times\) an odd integer, so it is a multiple of \(4\) but not of \(8\)

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