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.