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=\frac32\text{ or }n=-2`

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

Then for
`n=\frac32` we find
`2\left(\frac32\right)+1=4`, but this is not odd.

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