Download Homework Solution 5 (p. 331, 2, 4, 6, 7,10,13,17,19, 21, 23) 2. (a

Homework Solution 5 (p. 331, 2, 4, 6, 7,10,13,17,19, 21, 23)
2. (a) Solution: We solve 7x ≡ 1 (mod 19), or equivalently 7x+19y = 1. The
equation has integer solutions since gcd(7, 19)=1. We use the ”reverse Euclidean Alogirthm” method (see chapter 2 on how to solve linear Diophantine
equations). Write
19 = 2 × 7 + 5, 7 = 1 × 5 + 2, 5 = 2 × 2 + 1,
so
1 = 5−2×2 = 5−2×(7−1×5) = −2×7+3×5 = −2×7+3(19−2×7) = −8×7+2×19.
Hence 7−1 ≡ −8 ≡ 11 (mod 19).
(b) 39−1 ≡ 39 (mod 95).
(c) 91−1 ≡ 21 (mod 191).
4. (a) Since p ≡ 3 (mod 4), p = 3 + 4k, where k is an integer. So (p + 1)/4 =
1 + k which is an integer.
(p + 1)
(b) Since, by (a), (p + 1)/4 is an integer, and
×4=p+1≡1
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(mod p), by the definition, (p + 1)/4 is the multiplicative inverse of 4 modulo
p.
6. (a) To solve the equation 23x ≡ 1 (mod 91), it is equivalent to solve
linear Diophantine equation 23x + 91y = 1. Since gcd(23, 91) = 1, there is a
solution to 23x+91y = 1. We use the ”reverse Euclidean Alogirthm” method
(see chapter 2 on how to solve linear Diophantine equations). 91 = 4(23) − 1,
so x = 4. So the solution to 23x ≡ 1 (mod 91) is x ≡ 4 (mod 91).
(b) To solve 117x ≡ 136 (mod 241), it is equivalent to 117x + 241y = 136.
We then follow the steps outlined in chapter 2 in solving the Diophantine
equation (see my notes):
Step 1: Solve the equation 117x + 241y = 1 (where 1=d=gcd(117, 241))
using the reverse Euler algorithm: 241 = 2 × 117 + 7 117 = 16 × 7 + 5,
7 = 1×5+2, 5 = 2×2+1. So 1 = 5−2×2 = 5−2×(7−5) = −2×7+3×5 =
−2 × 7 + 3 × (117 − 16 × 7) = 3 × 117 − 50 × 7 = 3 × 117 − 50(241 − 2 × 117) =
103 × 117 − 50 × 241. So we get 103 × 117 − 50 × 241 = 1, i.e. x = 103 is the
solution (you don’t need to worry about y since we’ll modulo 241 anyway).
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Step 2: My multiplying 136 on both sides of 103 × 117 − 50 × 241 = 1, you
see that x0 = 103 × 136 is the solution to the equation 117x + 241y = 136.
By calculation, x0 = 14008 ≡ 30 (mod 241).
Hence, the general solution to 117x ≡ 136 (mod 241) is x ≡ 30 (mod
241).
Since gcd(117, 241) = 1, from the theorem in the chapter summary, the
equation 117x ≡ 136 (mod 241) has ONLY one solution modulo 241,
hence, the general solution to 117x ≡ 136 (mod 241) is x ≡ 30 (mod 241).
Check: 117 × 30 = 3510 ≡ 136 (mod 241). x ≡ 103 (mod 241).
(c) x ≡ 147 (mod 314).
7. (a) To solve 23x ≡ 7 (mod 91), we use 6(a) that x ≡ 4 (mod 91) is a
solution to 23x ≡ 1 (mod 91), i.e. 23 × 4 = 1 + 91k1 . Multiplying 7 on both
sides yields 23 × 28 = 7 + 91k2 . Hence 28 is a solution to 23x ≡ 7 (mod
91). Since gcd(23, 91)=1, the general (only solution mod 91) solution to the
equation 23x ≡ 7 (mod 91) is x ≡ 28(mod 91).
(b) x ≡ 30 (mod 241).
(c) x ≡ 40 (mod 314).
10. Since gcd (15, 108)=3—21, there are three solutions modulo 108. They
are: x ≡ 23 (mod 108), x ≡ 59 (mod 108), x ≡ 95 (mod 108),
13. Since gcd(102, 432)=2 does not divide 37, there is no solution to 102x ≡
37 (mod 432).
17. Since gcd(18, 156) = 6, there are either 0 solutions (when 6 does not
divide b) or exactly 6 solutions (when 6 | b) modulo 156.
19. The possible values of gcd(a, 42) are the divisors of 42. Thus the possible
number of solutions is 0 (for example when a = 0), 1 for example when
a = 1), 2(for example when a = 2), ,3 (for example when a = 3),6 (for
example when a = 6),7 (for example when a = 7),14 (for example when
a = 17),21 (for example when a = 21), and 42 (for example when a = 42).
21. (a) We need find a such that gcd(a, 78) = 13. Hence a = 13 and 65.
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(b) We need find a such that gcd(a, 78) = 26. Hence a = 26 and 52.
23. To solve the system of linear congruences of two variables 13x + 2y ≡ 1
(mod 15), 10x + 9y ≡ 8 (mod 15), the idea is (similar to the linear algebra
in solving linear equations) to eliminate one variable, say y. To do so, one
naturally wants to multiply 9 on the both sides of the first equation and
to multiply 2 on the second equation. Let’s see what happened after we
multiply 9 to the both sides of the first equation, we get 117x + 18y ≡ 9
(mod 135), NOT 117x + 18y ≡ 9 (mod 15). In other words, we can replace
the equation 13x + 2y ≡ 1 (mod 15) by 117x + 18y ≡ 9 (mod 135). However,
this is not our purpose, our purpose is to replace he equation 13x + 2y ≡ 1
(mod 15) by 117x + 18y ≡ 9 (mod 15). Let us see whether these equations
are equivalent (if they are equivalent, then we can certainly replace one with
another). Let’s see what is the meaning of 13x + 2y ≡ 1 (mod 15). The
meaning 13x + 2y =1 (mod 15) is that 15|(13x + 2y − 1). The meaning 117x
+ 18y =9 (mod 15) is that 15|(117x + 18y − 9). Write c = (13x + 2y − 1)
and note that 117x + 18y − 9 = 9c. So it gets down to see whether 15|c is
equivalent (if and only if) 15|9c. This is FALSE, since gcd(15, 9) 6= 1. (we
have the theorem that ”if gcd(a, b) = 1, then a|(bc) if and only if a|c”). So
multiplying 9 to the both sides of the first equation brings the TROUBLE.
Note that you can multiply 2 on the second equation (since 10x + 9y ≡ 8
(mod 15) is equivalent to 20x + 18y ≡ 16 (mod 15), due to the fact that
gcd(15, 2)=1).
So to avoid the mistake, the trick is to modify the first equation 13x+2y ≡
1 (mod 15). We subtract the second equation from the first equation to get
−3x + 7y ≡ 7 (mod 15), or 12x + 7y ≡ 7 (mod 15). Now we solve, instead
the two new equations: 13x + 2y ≡ 1 (mod 15), 12x + 7y ≡ 7 (mod 15) We
now CAN eliminate the variable y multiplying 7 to the first equation (note:
gcd(15, 7)=1) and multiplying 2 to the second equation (note: gcd(15, 2)=1)
to get 91x + 14y ≡ 7 (mod 15), 24x + 14y ≡ 14 (mod 15). Subtracting the
first from the second, we get 67x ≡ −7 (mod 15) or 7x ≡ 8 (mod 15). It
solution is x ≡ 14 (mod 15). Substituting this back to 13x + 2y ≡ 1 (mod
15), one gets 2y ≡ 14 (mod 15). Thus y ≡ 7 (mod 15).
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