Let [tex]2n+1[/tex] be the smaller integer. The larger integer is then [tex]2n+3[/tex], and we have
[tex](2n+1)^2=10+(2n+3)\implies4n^2+4n+1=2n+13[/tex]
[tex]\implies4n^2+2n-12=0[/tex]
[tex]\implies2n^2+n-6=0[/tex]
[tex]\implies(2n-3)(n+2)=0[/tex]
[tex]\implies 2n-3=0\text{ or }n+2=0[/tex]
[tex]\implies n=\dfrac32\text{ or }n=-2[/tex]
We omit [tex]n=-2[/tex], since [tex]2(-2)+1=-3[/tex] is negative.
Then for [tex]n=\dfrac32[/tex] we find [tex]2\left(\dfrac32\right)+1=4[/tex], but this is not odd.
There are no consecutive odd integers that satisfy the given condition!
