Intro to Proofs: Greatest Common Divisors

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The discussion focuses on solving two problems related to the greatest common divisor (gcd) of integers a and b. In part (a), it is established that if gcd(a, b) = d and ma + nb = d, then the integer solutions for xa + yb = d can be expressed in terms of k, where x and y are defined as x = m + k(b/d) and y = n - k(a/d). Part (b) explores the condition for the equation xa + yb = c to have integer solutions, demonstrating that such solutions exist if and only if d divides c. The participant expresses uncertainty about proving the "if and only if" condition, considering both direct and contrapositive approaches. The conversation highlights the challenges of understanding proofs in the context of gcd and integer solutions.
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Homework Statement



(a) Let a and b be integers with gcd(a,b)=d, and assume that ma+nb=d for integers m and n. Show that the solutions in Z to

xa+yb=d

are exactly

x=m+k(b/d), y=n-k(a/d)

where k∈Z.

(b) Let a and b be integers with gcd(a,b)=d. Show that the equation

xa+yb=c

has a solution (x,y)∈ ZxZ if and only if d|c.

Z refers to integers.

Homework Equations





The Attempt at a Solution



Ok so I'm pretty sure I figured out part (a).

Part (b) on the other hand is causing me problems. I know since it's an "if and only if" question, I have to prove it both ways.

This is what I have so far.

So we have some x and y, such as

x0a+y0b=d

Then we multiply both sides by some arbitrary element, such as n. Giving us,

(nx0)a+(ny0)b=nd=c

So that's the first part. Now I'm supposed to go the other way with it, showing that
xa+yb=c -> c=nd, n∈ Z.
 
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How about the contrapositive for the other direction? That might be easier.
 
So maybe this?

Suppose d∤c...

Then c≠nd

In which case, c≠nd≠(nx0)a+(ny0)b.

So like that? Or am I way off? Sorry I'm not very good at this... The teacher explains things way over everyone's head.
 
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