MATH THE EUCLIDEAN ALGORITHM

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1 MATH THE EUCLIDEAN ALGORITHM DR. ZACHARY SCHERR During our foray into number theory we learned about the greatest common divisor of two integers a and b (not both 0) and the importance of the existence of integral solutions to the equation ax + by = (a, b). The one element of this discussion which is missing is a way to compute the greatest common divisor as well as compute solutions to the above equation. The importance of being able to compute the x and y manifests itself in modular arithmetic in the following way. If (a, n) = 1 then any solution to ax + ny = 1 gives ax 1 (mod n) so that x is a multiplicative inverse of a modulo n. Thus in order to actually compute multiplicative inverses we must now how to solve ax + by = (a, b). The ey insight into beginning to understand how to compute lies in the following observation. Lemma 0.1. Suppose a, b, q, r Z with a = bq + r. Then (a, b) = (b, r). Proof. Let d = (a, b) and e = (b, r). Then since d a and d b we also get d a bq = r. Thus d e since e is the greatest common divisor of b and r. Similarly, since e b and e r we also get e bq + r = a so that e d and therefore d = e. Why should Lemma 0.1 be useful to us? Intuitively, one can reason as follows. If a > b are positive integers then we now we can find q, r Z with 0 r < b so that a = bq + r. Then Lemma 0.1 says that to compute (a, b) we can instead compute (b, r). We now that b < a and r < b so somehow the problem of computing (b, r) should be easier. We can turn this idea into an iterative and efficient algorithm. Given a > b let r 1 = a and r 0 = b. For i 0, as long as r i > 0, we define q i+1 and r i+1 via the division algorithm where 0 r i+1 < r i. We have r i 1 = r i q i+1 + r i+1 1

2 2 DR. ZACHARY SCHERR Proposition 0.2. There exists an n 0 so that r n > 0 and r n+1 = 0. For this n we have r n = (a, b). Proof. Let s first prove that r n exists. To do so, let S = {r i r i > 0}. Then S is a non-empty subset of the positive integers so it has a smallest element r n. Then r n 1 = r n q n+1 + r n+1 where 0 r n+1 < r n. Since r n is the smallest element of S we must have r n+1 = 0. By definition of the r i s, Lemma 0.1 tells us that (r i, r i+1 ) = (r i+1, r i+2 ). We can thus use induction to prove that (a, b) = (r 1, r 0 ) = (r n, r n+1 ). Of course r n+1 = 0 so (r n, r n+1 ) = r n and the proof is complete. To not get lost in all the symbols, let s compute an example. Example 0.3. Let s compute (6540, 1206). We have so that (6540, 1206) = = = = = = = = It s worth pointing out that this algorithm translates into repeated modular arithmetic. We can rephrase Example 0.3 by saying (mod 1206) (mod 510) (mod 186) (mod 138) (mod 48) 48 6 (mod 42) 42 0 (mod 6). Now we d lie to be able to use this idea, the Euclidean Algorithm, to produce solutions to ax + by = (a, b). We can do this via forward or bacward substitution. Let s loo at a simple example.

3 MATH THE EUCLIDEAN ALGORITHM 3 Example 0.4. Let s consider 13 and 5. We have 13 = = = = Suppose now we wanted to solve 13x+5y = 1. We can use Example 0.4 to solve this. Foward substution would start with the first line of the Euclidean Algorithm, which says = 3 and substitute this equation into the second line of the Euclidean Algorithm to get 5 = (13 5 2) or equivalently = 2. We can continue to substitute into the third line to get which rewrites as = ( ) = 1. This gives us a solution to 13x + 5y = 1! Moreover, one can see that this method will wor much more generally. It is called forward substitution. We can also use bacward substitution which is a similar idea but starts at the second to last step of the algorithm and uses the previous line, to substitute. This gives or 3 2 = = 2 3 (5 3) = = 1. Then again we go to the previous line which says so substitute: or = 3 (13 5 2) 2 5 = = 1. Again to connect bac with modular arithmetic, we see for example that (mod 13)

4 4 DR. ZACHARY SCHERR and since 5 8 (mod 13) we get that the multiplicative inverse of 5 modulo 13 is 8. That is, in J 13 we have [5][8] = [1]. There exist faster ways to solve ax+by = (a, b) but almost all of them rely on first performing the Euclidean Algorithm. My favorite method, which I will include without proof (the proof is not hard and might be worth thining about) is the following. Let q 1, q 2,..., q n+1 be the sequence of quotients showing up in the Euclidean Algorithm. In Example 0.3 this sequence would be 5, 2, 2, 1, 2, 1, 7. Let P 1 = 0, P 0 = 1, Q 1 = 1 and Q 0 = 0. For i 1 we inductively define P i = q i P i 1 + P i 2, Q i = q i Q i 1 + Q i 2. For Example 0.3, after organizing into a table, we get so for example P 3 = 27 and Q 5 = 19. One can chec that on this table we always have P i Q i+1 P i+1 Q i = ±1. For example, = 1. at the very end of the table we have = 1. Notice that = 1206 and = Thus if we multiply both sides of the equation by 6 we get = 6 a solution to 1206x y = 6. Remarably, this method always wors. In Example 0.4, the quotients are 2, 1, 1, 2. This gives the table and sure enough = 1 so that = 1. It will tae a little bit of wor, but from the definition of the P i and Q i, one can figure out and prove why this table always gives the right answer.

5 MATH THE EUCLIDEAN ALGORITHM 5 I d lie to mae a few concluding and interesting remars regarding computation of the GCD. Herstein tells us that if a = p e 1 1 pe and b = p f 1 1 pf are prime factorizations for a and b then (a, b) = p g 1 1 pg where g i = min(e i, f i ). This gives us another way of computing GCDs, but in practice it is useless compared to the Euclidean Algorithm. The reason is because factoring integers is HARD. Try factoring 6540 and 1206 by hand. You might quicly realize that both are divisible by 6 and so 6540 = , 1206 = but then how easy is it to factor 1090 and 201? Even worse, suppose a and b are huge numbers with hundred of digits. It should be much easier to do repeated division than it is to factor. In fact, it is currently a giant open question in mathematics and computer science as to whether one can factor in a reasonable amount of time. Without bacground in computer science it s hard to define what a reasonable amount of time means, but intuitively the question centers around whether one can factor a number n in a more efficient manner than just trial and error of dividing it by integers from 2 up to n. Most people thin the answer is no you cannot, and this is the basis for much of modern cryptography.