(a) Use the Euclidean Algorithm to nd gcd (2017; 271) and use this to nd integers x and y so that gcd(2017; 271) = 2017x + 271y.

(b) Is it true that for all integers a and b, if not both a and b are zeros then not both 5a+3band3a+2barezeros,andgcd(5a+3b;3a+2b)=gcd(a;b)? Proveyouranswer. (c) Is it true that for all integers a and b, if not both a and b are zeros then not both 3a+4band3a+2barezeros,andgcd(3a+4b;3a+2b)=gcd(a;b)? Proveyouranswer.

(a) To find the gcd (2017, 271) using the Euclidean Algorithm, we start by dividing 2017 by 271:

2017 ÷ 271 = 7 remainder 134

Next, we divide 271 by the remainder we obtained:

271 ÷ 134 = 2 remainder 3

Continuing with the algorithm, we divide the previous remainder by the current remainder:

134 ÷ 3 = 44 remainder 2

Finally, we divide the previous remainder by the current remainder:

3 ÷ 2 = 1 remainder 1

Since we obtained a remainder of 1 in the previous step, the gcd (2017, 271) is the last nonzero remainder we obtained, which is 1.

Now, we can find integers x and y such that gcd (2017, 271) = 2017x + 271y. We will use the Extended Euclidean Algorithm to determine these integers:

Starting with the last two equations:

3 = 134 - 44(3)
2 = 44 - 2(134 - 44(3))
1 = -(134 - 44(3)) + 2(44)

Simplifying:

1 = 2(44) - 134 + 44(3)
1 = 2(44) + 44(3) - 134
1 = 2(2017) + 271(-5)

Thus, for gcd(2017, 271) = 1, the integers x = 2 and y = -5 satisfy the equation 2017x + 271y = gcd(2017, 271).

(b) To prove whether or not the statement is true, we can consider cases where both 5a + 3b = 0 and 3a + 2b = 0.

1. If both 5a + 3b = 0 and 3a + 2b = 0:
In this case, both a and b must be zero.
gcd(5a + 3b, 3a + 2b) = gcd(0, 0) = 0.
gcd(a, b) = gcd(0, 0) = 0.
Therefore, both statements are true.

2. If not both 5a + 3b = 0 and 3a + 2b = 0:
In this case, at least one of the equations is nonzero, meaning that a and b cannot both be zero.
gcd(5a + 3b, 3a + 2b) = gcd(a, b).
Therefore, the statement holds true for all integers a and b.

(c) Similarly, let's consider cases where both 3a + 4b = 0 and 3a + 2b = 0.

1. If both 3a + 4b = 0 and 3a + 2b = 0:
In this case, both a and b must be zero.
gcd(3a + 4b, 3a + 2b) = gcd(0, 0) = 0.
gcd(a, b) = gcd(0, 0) = 0.
Therefore, both statements are true.

2. If not both 3a + 4b = 0 and 3a + 2b = 0:
In this case, at least one of the equations is nonzero, which implies that a and b cannot both be zero.
gcd(3a + 4b, 3a + 2b) = gcd(a, b).
Therefore, the statement holds true for all integers a and b.