Question

Find two successive natural numbers if the square of the sum of those numbers exceeds the sum of their squares by $112$.

Answer 1

Let the two natural numbers be

$n$ and $n+1$

Square of the sum of those numbers $=(n+n+1{)}^2=(2n+1{)}^2$

Sum of their squares $=n^2+(n+1{)}^2$

Therefore from the given data we can incur that...

$(2n+1{)}^2=n^2+(n+1{)}^2+112$

$4n^2+4n+1=n^2+n^2+2n+1+112$

Rearrange the expression into a quadratic equation

$2n^2+2n-112=0$

$2n^2+16n-14n-112=0$

$2n(n+8)-14(n+8)=0$

$(2n-14)(n+8)=0$

From this we can say $n$ is $7$ or $-8$, but since n is a natural number, it cannot be $-8$.

Hence, $n$ is $7$, and $n+1=8$.

The numbers are $7$ and $8$, this is the required solution