Question

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

Answer 1

Let three consequenture natural numbers are n-1, n, and n+1

Now

Given $(n{)}^2=(n+1{)}^2-(n-1{)}^2+60$

$n^2=n^2+1+2n-n^2-1+2n+60$

$n^2-4n-60=0$

$n^2-10n+6n-60=0$

$n(n-10)+6(n-10)=0$

$(n+6)(n-10)=0$

[n=-6] and [n=10]

Hence three numbers are -7,-6,-5 or 9,10 and 11

But numbers $\ne -7,-6,-5$ then three numbers are 9,10 & 11