n is a positive integer. Show that, for all n, $(5n+3)^2-(5n-3)^2$ is a multiple of 4. Answer ...

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Worked solution

1. Apply the difference of two squares identity.

Let $a=5n+3$ and $b=5n-3$ .

$(5n+3)^2-(5n-3)^2=[(5n+3)+(5n-3)][(5n+3)-(5n-3)]$

2. Simplify and show the result is divisible by 4.

$[(5n+3)+(5n-3)][(5n+3)-(5n-3)]$
$=(10n)(6)$
$=60n$
$=4(15n)$

Since $n$ is a positive integer, $15n$ is an integer. Hence $4(15n)$ is a multiple of $4$ .

Answer

$(5n+3)^2-(5n-3)^2=60n=4(15n)$
Since $15n$ is an integer, $(5n+3)^2-(5n-3)^2$ is a multiple of $4$ .

AI analysis — may contain mistakes.

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