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Question from Anita:

Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the square of the other two by 60. Find the numbers.

Hi Anita,

To get started I would let the three consecutive natural numbers be $n-1, n, \mbox{ and } n+1.$ In this case the difference of the squares of the largest and smallest of the three natural numbers is

\[\left(n+1\right)^2 - \left(n-1\right)^{2}.\]

You could expand and simplify this expression but I would rather expand it as a difference of squares and then simplify it. Be careful with the signs.

Use the fact that "the middle number exceeds the difference of the square of the other two by 60" to create en equation and solve for $n.$ Make sure you verify your answer.

Penny

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