Question

Find two positive consecutive odd integers such that square of the smaller integer is 10 more than the larger integer

Answer 1

Let $2n+1$ be the smaller integer. The larger integer is then $2n+3$, and we have

$(2n+1)^2=10+(2n+3)\implies4n^2+4n+1=2n+13$

$\implies4n^2+2n-12=0$

$\implies2n^2+n-6=0$

$\implies(2n-3)(n+2)=0$

$\implies 2n-3=0\text{ or }n+2=0$

$\implies n=\dfrac32\text{ or }n=-2$

We omit $n=-2$, since $2(-2)+1=-3$ is negative.

Then for $n=\dfrac32$ we find $2\left(\dfrac32\right)+1=4$, but this is not odd.

There are no consecutive odd integers that satisfy the given condition!