Question

The sum of squares of two consecutive odd numbers is $202$. Find the numbers.

Answer 1

Let the two consecutive odd numbers are $n$ and $n+2$

Sum of their squares is $202$

So,

$n^2+(n+2{)}^2=202$

$n^2+n^2+4+4n=202$

$2n^2+4n-198=0$

$n^2+2n-99=0$

$n^2+11n-9n-99=0$

$n(n+11)-9(n+11)=0$

$(n-9)(n+11)=0$

$n=9$ or $n=-11$

For $n=9$,

the two numbers are $9$ and $11$

For $n=-11$

the two numbers are $-11$ and $-9$

So, the numbers are $9,11$ or $-11,-9$