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§ 1.2 Primes: Solved problems

23. (a) Use the Euclidean algorithm to find gcd (1776,1492).

Solution: We have

1776 = 1492 · 1 + 284;
1492 = 284 · 5 + 72;
284 = 72 · 3 + 68;
72 = 68 · 1 + 4;
68 = 4 · 17 .

Thus gcd (1776,1492) = 4.

(b) Use the prime factorizations of 1492 and 1776 to find gcd (1776,1492).

Solution: Since 1776 = 24 · 3 · 37 and 1492 = 22 · 373, Proposition 1.2.9 shows that gcd (1776,1492) = 22.

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24. (a) Use the Euclidean algorithm to find gcd (1274,1089).

Solution: We have

1274 = 1089 · 1 + 185;
1089 = 185 · 5 + 164;
185 = 164 · 1 + 21;
164 = 21 · 7 + 17;
21 = 17 · 1 + 4;
17 = 4 · 4 + 1.

Thus gcd (1274,1089) = 1.

(b) Use the prime factorizations of 1274 and 1089 to find gcd (1274,1089).

Solution: Since 1274 = 2 · 72 · 13 and 1089 = 32 · 112, we see that 1274 and 1089 are relatively prime.

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25. Give the lattice diagram of all divisors of 250. Do the same for 484.

Solution: The prime factorizations are 250 = 2 · 53 and 484 = 22 · 112. In each diagram, we need to use one axis for each prime. Then we can just divide (successively) by the prime, to give the factors along the corresponding axis. For example, dividing 250 by 5 produces 50, 10, and 2, in succession. These numbers go along one axis of the rectangular diagram.

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26. Find all integer solutions of the equation xy + 2y - 3x = 25.

Solution: If we had a product, we could use the prime factorization theorem. That motivates one possible method of solution.

xy + 2y - 3x = 25
(x + 2)y - 3x = 25
(x + 2)y - 3x -6 = 25 -6
(x + 2)y - 3(x + 2) = 19
(x + 2)(y - 3) = 19

Now since 19 is prime, the only way it can be factored is to have 1 · 19 = 19 or (-1) · (-19) = 19. Therefore we have 4 possibilities:

x+2 = 1,     x+2 = -1,     x+2 = 19,     or     x+2 = -19.

For each of these values there is a corresponding value for y, since the complementary factor must be equal to y - 3. Listing the solutions as ordered pairs (x,y), we have the four solutions

(-1,22),   (-3,-16),   (17,4), and   (-21,2).

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27. For positive integers a,b, prove that gcd (a,b) = 1 if and only if gcd(a2,b2) = 1.

Solution: Proposition 1.2.3 (d) states that
gcd (a,bc) = 1 if and only if gcd (a,b) = 1 and gcd (a,c) = 1. Using c = b gives
gcd (a,b2) = 1 if and only if gcd (a,b) = 1. Then a similar argument yields
gcd (a2,b2) = 1 if and only if gcd (a,b2) = 1.

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28. Prove that n - 1 and 2n - 1 are relatively prime, for all integers n > 1.
Is the same true for 2n - 1 and 3n - 1?

Solution: We can write (1)(2n-1) + (-2)(n-1) = 1, which proves that gcd (2n-1,n-1) = 1.

Similarly, (2)(3n-1) + (-3)(2n-1) = 1, and so gcd (3n-1,2n-1) = 1.

Comment: Is this really a proof? Yes--producing the necessary linear combinations is enough; you don't have to explain how you found them.

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29. Let m and n be positive integers. Prove that
gcd (2m - 1, 2n - 1) = 1 if and only if gcd (m,n) = 1.

Comment: We need to do the proof in two parts. First, we will prove that
if gcd (m,n) = 1, then gcd (2m - 1, 2n - 1) = 1. Then we will prove the converse, which states that
if gcd (2m - 1, 2n - 1) = 1, then gcd (m,n) = 1. To prove the converse, we will use a proof by contradiction, assuming that gcd (m,n) 1 and showing that this forces
gcd (2m - 1, 2n - 1) 1.
Before beginning the proof, we recall that the following identity holds for all values of x:

xk - 1 = (x - 1)(xk-1 + xk-2 + · · · + x + 1).

Solution: If gcd (m,n) = 1, then there exist a,b in Z with am+bn=1. Substituting x = 2m and k = a in the identity given above shows that 2m - 1 is a factor of 2am - 1, say
2am - 1 = (2m - 1)(s), for some s in Z. The same argument shows that we can write
2bn - 1 = (2n - 1)(t), for some t in Z. The proof now involves what may look like a trick (but it is a useful one). We have

1 = 21 - 1
    = 2am+bn - 2bn + 2bn - 1
    = 2bn(2am - 1) + 2bn - 1
    = 2bn(s)(2m - 1) + (t)(2n - 1)

and so we have found a linear combination of 2m - 1 and 2n - 1 that equals 1, which proves that
gcd (2m - 1, 2n - 1) = 1. If gcd (m,n) 1, say gcd (m,n) = d, then there exist p,q in Z with m = dq and n = dp. But then an argument similar to the one given for the first part shows that 2d - 1 is a common divisor of 2dq - 1 and 2dp - 1. Therefore
gcd (2m - 1, 2n - 1) 1, and this completes the proof.

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30. Prove that gcd (2n2 + 4n - 3, 2n2 + 6n - 4 ) = 1, for all integers n > 1.

Solution: We can use the Euclidean algorithm. Long division of polynomials shows that dividing
2n2 + 6n - 4 by 2n2 + 4n - 3 gives a quotient of 1 and a remainder of 2n - 1. The next step is to divide
2n2 + 4n - 3 by 2n - 1, and this gives a quotient of n + 2 and a remainder of n - 1. We have shown that

gcd (2n2 + 6n - 4, 2n2 + 4n - 3)
= gcd (2n2 + 4n - 3, 2n - 1)
= gcd (2n - 1, n - 1)

and so we can use Problem 1.2.28 (in this section) to conclude that
2n2 + 4n - 3 and 2n2 + 6n - 4 are relatively prime since 2n - 1 and n - 1 are relatively prime.

Comment: You could also continue with the Euclidean algorithm, getting
gcd (2n - 1, n - 1) = gcd (n-2, 1) = 1.


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