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Theorem 5.6 For any integers a and b not both zero, the Diophantine equation ax + by gcf(a, b) = has solutions. Partial proof: If a = 0, then ax + by gcf(a, b) reduces to by= b). Then (x, y) = (0, 1) or (x, y) = (0, -1) is a solution. Similarly, if b= 0, the equation has solutions. Since gef(a, b) = gef(a, b), the equation ax + by gcf(a, b) has integer solutions if and only if the equation ax + bly = gcf(a, b) has integer solu- tions. So we can assume a > 0 and b>0. Since gef(a, b) = gef(b, a), we can assume a > b. The idea of the proof is to reverse the steps in the Euclidean Algo- rithm for computing gef(a, b). For example, suppose that for two positive integers a and b with a >b, the Euclidean Algorithm stops after four applications of the Division Algorithm. That is, suppose and bq₁ + ₁ b = 11₁92 + 1₂ 7293 + 1₂ 7₂ + 0.

Theorem 5.6 For any integers a and b not both zero, the Diophantine equation ax + by gcf(a, b) = has solutions. Partial proof: If a = 0, then ax + by gcf(a, b) reduces to by= b). Then (x, y) = (0, 1) or (x, y) = (0, -1) is a solution. Similarly, if b= 0, the equation has solutions. Since gef(a, b) = gef(a, b), the equation ax + by gcf(a, b) has integer solutions if and only if the equation ax + bly = gcf(a, b) has integer solu- tions. So we can assume a > 0 and b>0. Since gef(a, b) = gef(b, a), we can assume a > b. The idea of the proof is to reverse the steps in the Euclidean Algo- rithm for computing gef(a, b). For example, suppose that for two positive integers a and b with a >b, the Euclidean Algorithm stops after four applications of the Division Algorithm. That is, suppose and bq₁ + ₁ b = 11₁92 + 1₂ 7293 + 1₂ 7₂ + 0.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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problem 7

7. Complete the proof of Theorem 5.6 by following the plan suggested by the partial proof in this section.
Transcribed Image Text:7. Complete the proof of Theorem 5.6 by following the plan suggested by the partial proof in this section.
Theorem 5.6 For any integers a and b not both zero, the Diophantine equation ax + by gcf(a, b) = has solutions. Partial proof: If a = 0, then ax + by gcf(a, b) reduces to by= b). Then (x, y) = (0, 1) or (x, y) = (0, -1) is a solution. Similarly, if b= 0, the equation has solutions. Since gef(a, b) = gef(a, b), the equation ax + by = gef(a, b) has integer solutions if and only if the equation lalx + [bly = gcf(a, b) has integer solu- tions. So we can assume a > 0 and b>0. Since gef(a, b) = gef(b, a), we can assume a > b. The idea of the proof is to reverse the steps in the Euclidean Algo- rithm for computing gef(a, b). For example, suppose that for two positive integers a and b with a >b, the Euclidean Algorithm stops after four applications of the Division Algorithm. That is, suppose bq₁ + ₁ b == 1192 + 1₂ r₂ = 1293 + 1₂ and ₂7394 +0, where all the q, and r are integers, and 0 <+1 <r for all i. Then gcf(a, b) = 13₁-1293 =41- (b-192)93 = a - bq₁bg + (a - bq₁)9293 = a(1 + 9293) + b(-9₁ 93 − 919293). Thus, gef(a,b) = ax + by, where x = 1 +929, and y=-(9₁ +93 +919293). Since all the q, are integers, we have found a solution to ax + by gcf(a, b). Notice that the last three steps of the Euclidean Algorithm scheme for gcf(a, b) are precisely the Euclidean Algorithm scheme for gcf(b, r). Thus our calculation shows that = 1 +9293. gcf(b, r₁) = v= bx' + r₁y' where x' = -q, and y' Now, by using (*) and the first step of the scheme, we obtain gef(a, b) = gef(b, r₁) = v= bx' + (a - bq₁)y' = ay' + b(x' - q₁y') = ax + by. Thus, we have not only proved the theorem for the case in which the Euclidean Algorithm terminates after four applications of the Division Algorithm, but also we have shown how to prove the theorem for this case from the case in which it Unit 5.2 1 Divisibility Properties of the Integers 215 terminates in three applications of the Division Algorithm. This sets the stage for a general proof by mathematical induction. You are asked to supply the details of the proof in Problem 7. We now use Theorem 5.6 to derive a very basic divisibility result.
Transcribed Image Text:Theorem 5.6 For any integers a and b not both zero, the Diophantine equation ax + by gcf(a, b) = has solutions. Partial proof: If a = 0, then ax + by gcf(a, b) reduces to by= b). Then (x, y) = (0, 1) or (x, y) = (0, -1) is a solution. Similarly, if b= 0, the equation has solutions. Since gef(a, b) = gef(a, b), the equation ax + by = gef(a, b) has integer solutions if and only if the equation lalx + [bly = gcf(a, b) has integer solu- tions. So we can assume a > 0 and b>0. Since gef(a, b) = gef(b, a), we can assume a > b. The idea of the proof is to reverse the steps in the Euclidean Algo- rithm for computing gef(a, b). For example, suppose that for two positive integers a and b with a >b, the Euclidean Algorithm stops after four applications of the Division Algorithm. That is, suppose bq₁ + ₁ b == 1192 + 1₂ r₂ = 1293 + 1₂ and ₂7394 +0, where all the q, and r are integers, and 0 <+1 <r for all i. Then gcf(a, b) = 13₁-1293 =41- (b-192)93 = a - bq₁bg + (a - bq₁)9293 = a(1 + 9293) + b(-9₁ 93 − 919293). Thus, gef(a,b) = ax + by, where x = 1 +929, and y=-(9₁ +93 +919293). Since all the q, are integers, we have found a solution to ax + by gcf(a, b). Notice that the last three steps of the Euclidean Algorithm scheme for gcf(a, b) are precisely the Euclidean Algorithm scheme for gcf(b, r). Thus our calculation shows that = 1 +9293. gcf(b, r₁) = v= bx' + r₁y' where x' = -q, and y' Now, by using (*) and the first step of the scheme, we obtain gef(a, b) = gef(b, r₁) = v= bx' + (a - bq₁)y' = ay' + b(x' - q₁y') = ax + by. Thus, we have not only proved the theorem for the case in which the Euclidean Algorithm terminates after four applications of the Division Algorithm, but also we have shown how to prove the theorem for this case from the case in which it Unit 5.2 1 Divisibility Properties of the Integers 215 terminates in three applications of the Division Algorithm. This sets the stage for a general proof by mathematical induction. You are asked to supply the details of the proof in Problem 7. We now use Theorem 5.6 to derive a very basic divisibility result.
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