also includes the proof of his Proposition 3). Let be relatively prime positive integers: without loss of generality is not itself prime, otherwise there is nothing to prove. Let therefore be a proper factor of , not necessarily prime: we wish to show that is a sum of two squares. Again, we lose nothing by assuming since the case is obvious. Let be non-negative integers such that are the closest multiples of (in absolute value) to respectively. Notice that the differences and are integers of absolute value strictly less than : indeed, when is even, gcd; otherwise since gcd, we would also have gcd. Multiplying out we obtain uniquely defining a non-negative integer . Since divides both ends of this equation sequence it follows that must also be divisible by : say . Let be the gcd of and which by the co-primeness of is relatively prime to . Thus divides , so writing , and , we obtain the expression for relatively prime and , and with , since Now finally, the descent step: if is not the sum of two squares, then by step (3.) there must be a factor say of which is not the sum of two squares. But and so repeating these steps (initially with in place of , and so on ad infinitum) we shall be able to find a strictly decreasing infinite sequence of positive integers which are not themselves the sums of two squares but which divide into a sum of two relatively prime squares. Since such an infinite descent is impossible, we conclude that must be expressible as a sum of two squares, as claimed. 5. Every prime of the form is a sum of two squares. (This is the main result of Euler's second paper). If , then by Fermat's Little Theorem each of the numbers is congruent to one modulo . The differences are therefore all divisible by . Each of these differences can be factored as Since is prime, it must divide one of the two factors. If in any of the cases it divides the first factor, then by the previous step we conclude that is itself a sum of two squares (since and differ by , they are relatively prime). So it is enough to show that cannot always divide the second factor. If it divides all differences , then it would divide all differences of