Explain why $(5n+2)^2-(5n-2)^2$ is a multiple of 8 for all integer values of n. Answer ...

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Worked solution

1. Recognise and apply the difference of two squares identity.

Let $a=5n+2$ and $b=5n-2$ .

$(5n+2)^2-(5n-2)^2=a^2-b^2=(a+b)(a-b).$

egin{aligned} (5n+2)^2-(5n-2)^2 &=[(5n+2)+(5n-2)][(5n+2)-(5n-2)]\\ &=(10n)(4). \end{aligned}

2. Rewrite the expression to show it is divisible by 8 for integer $n$.

$(10n)(4)=40n=8(5n).$

Since $n$ is an integer, $5n$ is also an integer. Hence $8(5n)$ is a multiple of 8.

Answer

$(5n+2)^2-(5n-2)^2=8(5n)=40n.$ Since $n$ is an integer, $5n$ is an integer, so $8(5n)$ is a multiple of 8.

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