BIG SQUARES!

$\begin{aligned} (a+1)^{2}-a^{2} & =a^{2}+2a+1-a^{2} \\ & =2a+1 \end{aligned}$

Putting $a=38918$

$\begin{aligned} \implies (38918+1)^{2}-38918^2 & =2.38918+1 \\ (38919)^{2}-38918^2 & =77836+1 \\ 38919^{2}-38918^2 & =\boxed{77837} \end{aligned}$

The whole point of this question is to use a calculator. The answer is 77837

The difference between squares of 2 consecutive integers is the sum of these 2 integers. Or $a^{2}-b^{2} = a+b$

Where a>b and their difference is 1

There is this own theorem of mine (which I am on the verge of proving) about the differences of squares of natural nos. The first part is that the difference always progresses in odd natural nos. Using that, you find only one odd no., 77837.

My solution: Notice that 38919 ends in 9 therefore when squared will end in 1. Notice that 38918 ends in 8 therefore when squared will end in 4. If 4 is subtracted from a number (greater than 1) that ends in 1, then the result ends in 7 out of which 77837 is the only option that ends in 7.