The difference of squares of two number is 88.

Solution:

Let the smaller numbers be $x$

Then according to question,

The larger number be $=2 x-5$, then

$(2 x-5)^{2}-x^{2}=88$

$4 x^{2}-20 x+25-x^{2}-88=0$

$3 x^{2}-20 x-63=0$

$3 x^{2}-20 x-63=0$

$3 x^{2}-27 x+7 x-63=0$

$3 x(x-9)+7(x-9)=0$

$(x-9)(3 x+7)=0$

$(x-9)=0$

$x=9$

Or

$(3 x+7)=0$

$x=\frac{-7}{3}$

Since, x being a positive integer so, x cannot be negative,

Therefore,

When $x=9$ then larger number be

$2 x-5=2 \times 9-5$

$=18-5$

 

$=13$

Thus, two consecutive number be either 9,13