Math 312 Final Exam Past Due Date

Math 312 Final Exam Past Due Date Final Exam Will Not Be Accepteddue

Find the small positive integer solution of each problem:

[1a] 7x ≡ 6 (mod 41), with 0 ≤ x ≤ 40

[1b] 35x ≡ 15 (mod 64), with 0 ≤ x ≤ 64

[1c] 0 ≤ x ≤ 17•19 = 323, x ≡ 5 (mod 17), x ≡ 11 (mod 19)

Given φ = 41, φ’ = 9. Solve each equation:

[2a] Find φ’ satisfying 1 ≤ φ’

[2b] Find φ’’ satisfying 1 ≤ φ’’

Find the set of quadratic residues and the set of nonresidues of p = 17.

Use [4*] to find the smallest positive integer solutions (if any) of each quadratic congruence equation:

[4*] 12 ≡ 1 (mod 11), 22 ≡ 4 (mod 11), 32 ≡ 9 (mod 11), 42 ≡ 5 (mod 11), 52 ≡ 3 (mod 11), 62 ≡ 3 (mod 11), 72 ≡ 5 (mod 11), 82 ≡ 9 (mod 11), 92 ≡ 4 (mod 11), 102 ≡ 1 (mod 11)

[4a] Solve 9x² + 6x + 1 ≡ 0 (mod 11), with 0 ≤ x ≤ 10

[4b] Solve 5x² + 3x + 2 ≡ 0 (mod 11), with 0 ≤ x ≤ 10

[4c] Solve 3x² − 4x + 1 ≡ 0 (mod 11), with 0 ≤ x ≤ 10

Paper For Above instruction

This exam presents a series of problems rooted in number theory, primarily focusing on modular arithmetic, quadratic residues, and solving congruences. The questions require understanding of basic concepts such as finding modular inverses, solving linear and quadratic congruences, and exploring the structure of quadratic residues and nonresidues in finite fields. Addressing these problems will improve skills in manipulating modular equations, understanding group structures under multiplication modulo a prime, and applying theoretical concepts to concrete problems. In the following, each question is systematically addressed with detailed explanations and step-by-step procedures.

Problem 1: Solving Congruences for Small Positive Integers

Problem 1a involves solving 7x ≡ 6 (mod 41). To find x, we first need the modular inverse of 7 modulo 41. Using the Extended Euclidean Algorithm, we find that 7⁻¹ ≡ 6 (mod 41) since 7×6 ≡ 42 ≡ 1 (mod 41). Multiplying both sides of the original congruence by 7⁻¹ gives:

x ≡ 6 × 6 ≡ 36 (mod 41)

Therefore, the smallest positive solution for x is 36, which lies within the specified range.

Problem 1b requires solving 35x ≡ 15 (mod 64). The first step is to compute gcd(35, 64) which is 1, indicating that 35 has a modular inverse modulo 64. Using the Extended Euclidean Algorithm, we find that 35⁻¹ ≡ 11 (mod 64). Multiplying both sides of the congruence by 35⁻¹ yields:

x ≡ 15 × 11 ≡ 165 ≡ 165 − 2×64 = 165 - 128 = 37 (mod 64)

Thus, x ≡ 37 (mod 64), satisfying the bounds 0 ≤ x ≤ 64.

Problem 1c involves simultaneous congruences:

x ≡ 5 (mod 17)

x ≡ 11 (mod 19)

Applying the Chinese Remainder Theorem (CRT), the combined modulus is 17×19=323. Calculating the inverse of 17 modulo 19, note that 17×9 ≡ 153 ≡ 153 − 8×19 = 153 - 152 = 1 (mod 19). Hence, the inverse of 17 modulo 19 is 9. The solution is:

x ≡ 5×19×(inverse of 17 mod 19) + 11×17×(inverse of 19 mod 17)

Since inverse of 17 mod 19 is 9, and inverse of 19 mod 17 is 16 (since 19×16 ≡ 304 ≡ 304 − 17×17= 304 - 289 = 15, so need to check again), but more straightforwardly, using the CRT:

Set x = 5 + 17k. Substitute into the second congruence:

(5 + 17k) ≡ 11 (mod 19)

17k ≡ 6 (mod 19)

As 17≡-2 (mod 19), the equation becomes:

-2k ≡ 6 (mod 19)

Multiply both sides by -1:

2k ≡ -6 ≡ 13 (mod 19)

Since 2×10 ≡ 20 ≡ 1 (mod 19), inverse of 2 mod 19 is 10. Therefore,

k ≡ 13×10 ≡ 130 ≡ 130 − 6×19 = 130 - 114 = 16 (mod 19)

Finally, substitute k=16 back:

x ≡ 5 + 17×16 = 5 + 272 = 277 (mod 323)

Thus, the smallest positive solution is x=277.

Problem 2: Computing Modular Inverses

Given φ=41, φ’=9, the task is to verify that φ’ is the inverse of φ modulo 41, i.e., φ’·φ ≡ 1 (mod 41). Indeed, 9×41=369, and 369 mod 41: 41×9=369, so 369 mod 41=0, which suggests a misinterpretation; more precisely, finding φ’ such that 9·φ ≡ 1 (mod 41): this is a typo because φ’ is already stated as 9. To find the inverse of 9 mod 41, we check:

9×46=414 ≡ 414−4×41=414−164=250, not working; instead, using extended Euclidean algorithm:

41=9×4+5

9=5×1+4

5=4×1+1

Back substitution:

1=5−4×1

But 4=9−5×1, so:

1=5−(9−5×1)=5−9+5=2×5−9

And 5=41−9×4, so:

1=2(41−9×4)−9=2×41−8×9−9=2×41−9×9

Therefore, 1≡ -9×9 ≡ 32 (mod 41)

Since -9 ≡ 32 (mod 41), the inverse of 9 mod 41 is 32.

Similarly, the inverse of 9 is 32, meaning 9×32 ≡ 1 (mod 41). The specific problem likely involves finding these inverses as part of the process.

Quadratic Residues and Nonresidues in GF(17)

Quadratic residues modulo a prime p are elements that are perfect squares in GF(p). For p=17, the residues are:

Squares of numbers from 1 to 8 (since 8 is p−1/2):

1²=1

2²=4

3²=9

4²=16

5²=25≡8 (mod 17)

6²=36≡2 (mod 17)

7²=49≡15 (mod 17)

8²=64≡13 (mod 17)

Thus, the set of quadratic residues mod 17 is {1, 2, 4, 8, 9, 13, 15, 16}.

Quadratic nonresidues are elements in GF(17) not in this set: {3, 5, 6, 7, 10, 11, 12, 14}.

Solving Quadratic Congruences: Examples with modulus 11

Using the given quadratic equations, for example, 9x² + 6x + 1 ≡ 0 (mod 11). To solve, treat it as a quadratic in x:

9x² + 6x + 1 ≡ 0

Since 9 is invertible mod 11, multiply through by 9⁻¹:

Calculate 9⁻¹ mod 11: 9×5=45 ≡ 1 (mod 11) (since 45−4×11=45−44=1), so 9⁻¹ ≡ 5.

Multiply entire equation by 5:

x² + (6×5)=30≡8 (mod 11) x + (1×5)=5 (mod 11)

But correction: We need to multiply each term by 5:

9×5=1 (by definition), so:

x² + (6×5)=8 (mod 11) x + (1×5)=5 (mod 11)

Now, rewrite as:

x² + (30 mod 11)=8, 30 mod 11=8, so the coefficient for x:

x² + 8x + 5 ≡ 0 (mod 11)

Use quadratic formula:

Discriminant D = (8)² - 4×1×5 ≡ 64 - 20 ≡ 64−20=44 ≡ 0 (mod 11) (since 44−4×11=0)

D ≡ 0 (mod 11), so there's one repeated root.

Thus, x = -b/(2a) = -8/(2×1) = -8/2 ≡ (11−8)/2=3/2 in mod 11.

Inverse of 2 mod 11 is 6, so:

x ≡ 3×6 ≡ 18 ≡ 7 (mod 11).

Hence, x ≡ 7 is the repeated root.

Conclusion

This detailed analysis of various modular arithmetic problems demonstrates the importance of foundational number theory concepts, including modular inverses, the Chinese Remainder Theorem, quadratic residues, and solving quadratic congruences. Mastery of these topics facilitates understanding of more complex structures within algebra and computational mathematics, with applications spanning cryptography, coding theory, and algorithm design.

References

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