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Decryption Stage While everyone can do the encryption using public (n,e), not everyone can decrypt. To do this, you need to know the exact p and q, which is very large and kept secret in practice. Actually, the secret key d used in decryption is an inverse of e mod (p - 1)(q-1), i.e., ed = 1 mod (p - 1)(q-1). According to the corollary given below, d can decrypt (m² mod n) back to m by another exponential operation. Note that m 1), e has an inverse d modulo x if and only ife and x are relatively prime. (Hint: For left to right, try run Euclidean Algorithm. For right to left, use Be'zout's Theorem.)

Decryption Stage While everyone can do the encryption using public (n,e), not everyone can decrypt. To do this, you need to know the exact p and q, which is very large and kept secret in practice. Actually, the secret key d used in decryption is an inverse of e mod (p - 1)(q-1), i.e., ed = 1 mod (p - 1)(q-1). According to the corollary given below, d can decrypt (m² mod n) back to m by another exponential operation. Note that m 1), e has an inverse d modulo x if and only ife and x are relatively prime. (Hint: For left to right, try run Euclidean Algorithm. For right to left, use Be'zout's Theorem.)

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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Decryption Stage While everyone can do the encryption using public (n,e), not everyone can decrypt. To do this, you need to know the exact p and q, which is very large and kept secret in practice. Actually, the secret key d used in decryption is an inverse of e mod (p-1)(q - 1), i.e., ed = 1 mod (p-1)(q-1). According to the corollary given below, d can decrypt (m² mod n) back to m by another exponential operation. Note that m <n so that m mod n = m. Corollary 1. Given the p, q, e, m as above. For d, if ed = 1 mod (p-1)(q-1), (m²) = m mod (pq). (Note: To prove this, you need to learn about Fermat's Little Theorem and Chinese Remainder Theorem. These two are core principles behind RSA encryption. Due to time limits, they are not included or needed in this worksheet. You may refer to other materials if interested.) Problem 4. Prove we can always find d. That is, generally for any integer e and x (x > 1), e has an inverse d modulo x if and only if e and x are relatively prime. (Hint: For left to right, try run Euclidean Algorithm. For right to left, use Be'zout's Theorem.)
Transcribed Image Text:Decryption Stage While everyone can do the encryption using public (n,e), not everyone can decrypt. To do this, you need to know the exact p and q, which is very large and kept secret in practice. Actually, the secret key d used in decryption is an inverse of e mod (p-1)(q - 1), i.e., ed = 1 mod (p-1)(q-1). According to the corollary given below, d can decrypt (m² mod n) back to m by another exponential operation. Note that m <n so that m mod n = m. Corollary 1. Given the p, q, e, m as above. For d, if ed = 1 mod (p-1)(q-1), (m²) = m mod (pq). (Note: To prove this, you need to learn about Fermat's Little Theorem and Chinese Remainder Theorem. These two are core principles behind RSA encryption. Due to time limits, they are not included or needed in this worksheet. You may refer to other materials if interested.) Problem 4. Prove we can always find d. That is, generally for any integer e and x (x > 1), e has an inverse d modulo x if and only if e and x are relatively prime. (Hint: For left to right, try run Euclidean Algorithm. For right to left, use Be'zout's Theorem.)
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