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I have my problem here. I can't solve it on my own. I need a solution to prove my answer. Does anybody know this? thanks

"Find the three consecutive numbers that have the property that the square of the middle number is greater by 1 than the product of the other two numbers."

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    $\begingroup$ Hint: You have three consecutive numbers $n-1$, $n$, and $n+1$. This isn't abstract algebra, by the way. $\endgroup$ – Lost Feb 14 '14 at 7:26
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Let the numbers be $n,n+1,n+2$. $$(n+1)^2 = n(n+2)+1$$ $$(n+1)^2 = n^2+2n+1$$ This is always true, isn't it ?

So you do not have a unique triplet of $3$ such consecutive numbers.!!

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First step is to formalize the property. If $n$ is the middle of the three numbers, the numbers are $n-1,n,n+1$ and the problem reads:

Find $n$ such that $n^2=(n+1)(n-1)+1$

Hint: $n$ is not uniquely determined by this.

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Find the three consecutive numbers that have the property that the square of the middle number is greater by 1 than the product of the other two numbers.

First, we need to reconstruct the problem: we are asked to find three consecutive numbers ($n-1,n,n+1$) such that the square of the middle ($n^2$) is greater by $1$ than the product of the other two numbers: $n^2=(n-1)(n+1)+1$. So the problem is equivalent to solving that equation for $n$: $$n^2=(n-1)(n+1)+1 \iff n^2=n^2+n-n-1+1=n^2$$

So you can take any number $n$, square it, and it will be equal to the product of $n-1$ and $n+1$ plus $1$. Example: $1,2\rm\,and\,3$: $2^2=4=3\cdot1+1=4.$

I hope this helps.
Best wishes, $\mathcal H$akim.

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Any consecutive three numbers can be the answers as the multiplication of the first and third numbers would be 1 less than the square of the number in the middle.

$$(n-1)(n+1) = n^2 - 1$$

For example, 1, 2, 3 ==> 1*3 = 3 is 1 less than 2*2 = 4.

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