Question 116046
If one of the integers is n, then the other one is n + 2, and their squares are {{{n^2}}} and {{{(n+2)^2}}}.


Then the sum of the squares is:
{{{n^2+(n+2)^2=209}}}


Expand the binomial and simplify:
{{{n^2+n^2+4n+4=209}}}
{{{2n^2+4n+4=209}}}
{{{2n^2+4n-205=0}}}, and now you have a problem because this equation does not have rational roots, which means there is no solution to the problem as you stated it.


However, not being satisfied with that result, I did a little 'easter-egging' and discovered that {{{11^2+13^2=290}}} and concluded that you made a small typographical error when you typed the problem.


So, let's start over:


If one of the integers is n, then the other one is n + 2, and their squares are {{{n^2}}} and {{{(n+2)^2}}}.


Then the sum of the squares is:
{{{n^2+(n+2)^2=290}}}


Expand the binomial and simplify:
{{{n^2+n^2+4n+4=290}}}
{{{2n^2+4n+4=290}}}
{{{2n^2+4n-286=0}}}
{{{n^2+2n-143=0}}}
{{{(n+13)(n-11)=0}}}


So n = 11 or n = -13.  But we can eliminate the -13 result because we are looking for consecutive odd POSITIVE integers.  Therefore, the first integer is 11 and the second is 11 + 2, or 13.


Please be careful and review your input for accuracy in the future. 


Hope that helps,
John