Question 470223
Let {{{x}}} is the odd number, next odd number is {{{(x+2)}}}, odd number before {{{x}}} is {{{(x-2)}}}.
The sum of the squares of three consecutive odd numbers is {{{(x-2)^2+x^2+(x+2)^2=2531}}}
Use formula {{{(a+-b)^2=a^2+-2ab+b^2}}}
{{{x^2-4x+4+x^2+x^2+4x+4=2531}}}
{{{3x^2=2531-8}}}
{{{3x^2=2523}}}
{{{x^2=2523/3}}}
{{{x^2=841}}}
{{{x1=sqrt(841)}}} or {{{x2=-sqrt(841)}}}
{{{x1=29}}} or {{{x2=-29}}}
If {{{x=29}}} the number
The next odd number is {{{29+2=31}}}, the third odd number is {{{29-2=27}}}
If {{{x=-29}}} the number
The next odd number is {{{-29+2=-27}}}, the third odd number is {{{-29-2=-31}}}
Answer: (27,29,31) and (-31,-29,-27)