This is the solution sent in by Yatir Halevi. Thanks Yatir. A correct solution was also received from Andrei Lazanu.

Let's say we want to find the square of $a$ .

We know that $a^{2} = a^{2} - b^{2} + b^{2} = (a + b) \times (a - b) + b^{2}$

and for every a, we can pick a certain b that will make the calculation $a^{2}$ as easy as possible.

For instance if we take $a = 35$ , we can take $b = 5$ ,

we get $35^{2} = (35 + 5) \times (35 - 5) + 5^{2} = 40 \times 30 + 25 = 1200 + 25 = 1225$

So, if $a$ is a number that ends with a 5: it can be written as

$a = 10 \times q + 5$ $a^{2} = (10 q + 5)^{2} = (10 q + 5 - 5) \times (10 q + 5 + 5) + 25 = 10 q (10 q + 10) + 25 = 10^{2} q (q + 1) + 25$ .

So $a^{2}$ is equal to q(q+1) plus two zeros after it $(10^{2})$ that are "stolen" by the 25 that is added on.