The consecutive odd integers whose sum is  ${45}^2-{21}^2$ are

Let n consecutive odd integers are
2m + 1, 2m + 3, 2m + 5, ……, 2m + 2n – 1
Given that,
${45}^2-{21}^2=(2m+1)+(2m+3)+(2m+5)+\mathrm{.......}+(2m+2n-1)$$=2mn+(1+3+\mathrm{5.....}+(2n-1))=2mn+n^2=m^2+2mn+n^2-m^2$$\Rightarrow {45}^2-{21}^2={(m+n)}^2-m^2$

$\Rightarrow m+n=45\&m=21$

$\Rightarrow n=24\&m=21$

Hence, the numbers are

43, 45, ……, 89