Use the Chinese remainder theorem to solve: Suppose a and b are relatively prime positive integers. By

Use the Chinese remainder theorem to solve:

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Suppose a and b are relatively prime positive integers. By Theorem 4.2 we can choose a so that ar is in any congruence class we want modulo b. For that matter we can also choose y so that by is in any congruence class we want modulo a. To emphasize the symmetry between a and b we will change the notation and consider relatively prime positive integers b₁ and b₂. By Theorem 4.2 we can find ₁ such that b2a11 (mod by). Then b₂a1c1c₁ (mod by), where c₁ is completely arbitrary. Likewise we can find 2 such that b₁2 1 (mod b₂), and then b₁2202 = c₂ (mod b₂), where c₂ is arbitrary. The important implication of the above congruences is that we can make the sum z = b₂x1c₁+b₁2c2 simultaneously congruent to whatever we want modulo both by and b2. For z = b₂x1c₁ + b₁₂c₂=b₂x1c₁+0=₁ (mod b₁), and z = b₂101 + b₁2C₂ = 0 + b₁x₂0₂ = C₂ (mod b₂). 2= The ability to find an integer z such that 2=C₁ (mod b₁) and zc₂ (mod b₂), where c₁ and c2 are arbitrary, has many applications. C2 Suppose a and b are relatively prime positive integers. By Theorem 4.2 we can choose a so that ar is in any congruence class we want modulo b. For that matter we can also choose y so that by is in any congruence class we want modulo a. To emphasize the symmetry between a and b we will change the notation and consider relatively prime positive integers b₁ and b₂. By Theorem 4.2 we can find ₁ such that b2a11 (mod by). Then b₂a1c1c₁ (mod by), where c₁ is completely arbitrary. Likewise we can find 2 such that b₁2 1 (mod b₂), and then b₁2202 = c₂ (mod b₂), where c₂ is arbitrary. The important implication of the above congruences is that we can make the sum z = b₂x1c₁+b₁2c2 simultaneously congruent to whatever we want modulo both by and b2. For z = b₂x1c₁ + b₁₂c₂=b₂x1c₁+0=₁ (mod b₁), and z = b₂101 + b₁2C₂ = 0 + b₁x₂0₂ = C₂ (mod b₂). 2= The ability to find an integer z such that 2=C₁ (mod b₁) and zc₂ (mod b₂), where c₁ and c2 are arbitrary, has many applications. C2

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To solve the given congruences using the Chinese Remainder Theorem we have the f... View the full answer