In the exercises that follow the chapter on algebraic manipulations, there is a question that pertains to expressing an integer as the sum of two other integers squared. We are then asked to find expressions for the multiples of , namely and . In this post, we’ll take a look at the solution provided by the authors for the first half of the question, , and then come up with a method of our own to solve the second half, .

Question

If is an integer that can be expressed as the sum of two integer squares, show that both and can also be expressed as the sum of two integer squares.

Solution

From the question, since it is the sum of two integer squares. This means . We need to find two integers such that when their squares are summed, we end with . From the solution, these are the numbers (a + b) and (a - b) because when they are squared and summed, we get . This is the result of . Go ahead and expand them to verify the result.

What do we deduce from this? We find that both the expressions contributed an and a . These were added together to get the final result. How do we use this to get ? Notice that we’re squaring the integers. This means that, for example, one of them would have to contribute an and the other would have to contribute a ; similar logic applies for .

This leaves us with two pairs of numbers — and . Let’s square and sum both of these numbers one-by-one.