Question

Find two consecutive odd positive integers, sum of whose squares is 970.

Answer 1

Let the two consecutive odd positive integers be $n$ and $n+2$

So, according to question

$n^2+(n+2{)}^2=970$

$n^2+n^2+4n+4=970$

$2n^2+4n+4-970=0$

$2n^2+4n-966=0$

$n^2+2n-483=0$

$n^2+23n-21n-483=0$

$n(n+23)-21(n+23)=0$

$(n+23)(n-21)=0$

So,$n=21,-23$

$n$ can not be negative.

Consecutive odd positive integers are $21$ and $21+2=23$