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

Question

asked Aug 9, 2020 87.4k views

1 Answer

Scottt · Answer 1 · 2020-08-13T12:52:22+0000

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!