Question

4. Let a, b, c be integers.

(a) Prove if gcd(ab, c) = 1, then gcd(a, c) = 1 and gcd(b, c) = 1. (Hint: use the GCD characterization theorem.)

(b) Prove if gcd(a, c) = 1 and gcd(b, c) = 1, then gcd(ab, c) = 1. (Hint: you can use the GCD characterization theorem again but you may need to multiply equations.)

(c) You have now proved that “gcd(a, c) = 1 and gcd(b, c) = 1 if and only if gcd(ab, c) = 1.” Is this statement true if the greatest common divisor is not 1?

So is the following statement true: “gcd(a, c) = d and gcd(b, c) = d if and only if gcd(ab, c) = d?”

Answer #1

1. (a) Let a, b and c be positive integers. Prove that gcd(ac,
bc) = c x gcd(a, b). (Note that c gcd(a, b) means c times the
greatest common division of a and b)
(b) What is the greatest common divisor of a − 1 and a + 1?
(There are two different cases you need to consider.)

8. Let a, b be integers. (a) Prove or disprove: a|b ⇒ a ≤ b. (b)
Find a condition on a and/or b such that a|b ⇒ a ≤ b. Prove your
assertion! (c) Prove that if a, b are not both zero, and c is a
common divisor of a, b, then c ≤ gcd(a, b).

prove that if gcd(a,b)=1 then gcd (a-b,a+b,ab)=1

1.13. Let a1, a2, . . . , ak be integers with gcd(a1, a2, . . .
, ak) = 1, i.e., the largest
positive integer dividing all of a1, . . . , ak is 1. Prove that
the equation
a1u1 + a2u2 + · · · + akuk = 1
has a solution in integers u1, u2, . . . , uk. (Hint. Repeatedly
apply the extended Euclidean
algorithm, Theorem 1.11. You may find it easier to prove...

Prove that for all non-zero integers a and b, gcd(a, b) = 1 if
and only if gcd(a, b^2 ) = 1

Let a, b, c be natural numbers. We say that (a, b, c) is a
Pythagorean triple, if a2 + b2 = c2 . For example, (3, 4, 5) is a
Pythagorean triple. For the next exercises, assume that (a, b, c)
is a Pythagorean triple.
(c) Prove that 4|ab Hint: use the previous result, and a proof
by con- tradiction.
(d) Prove that 3|ab. Hint: use a proof by contradiction.
(e) Prove that 12 |ab. Hint : Use the...

Given that the gcd(a, m) =1 and gcd(b, m) = 1. Prove that gcd(ab,
m) =1

The greatest common divisor c, of a and b, denoted as c = gcd(a,
b), is the largest number that divides both a and b. One way to
write c is as a linear combination of a and b. Then c is the
smallest natural number such that c = ax+by for x, y ∈ N. We say
that a and b are relatively prime iff gcd(a, b) = 1. Prove that a
and n are relatively prime if and...

Prove that if a|n and b|n and gcd(a,b) = 1 then ab|n.

Let gcd(m1,m2) = 1. Prove that a ≡ b (mod m1) and a ≡ b (mod m2)
if and only if (meaning prove both ways) a ≡ b (mod m1m2). Hint: If
a | bc and a is relatively prime to to b then a | c.

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