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Any explanation on how to solve this would be greatly appreciated. Thank you.

True or False? Justify your answer.

Let a and b be two natural numbers.

(A) If (a,b)=d and sa+tb=d then (s,t)=1

(B) If (a, b) = d, then the equation a x + b y = d has one and only one solution x, y

for b)i know that it relates to this proposition: Let x0,y0 be a solution of ax + by = c. Then the general solution of ax+by = c is of the form x = x0 +z,y = y0 +w, where z,w is any solution of ax+by=0.

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    Hint for (A): Consider $a/d$ and $b/d$. – 2017-01-25
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    A: Both $sa$ and $tb$ are multiples of $(a,b)(s,t)$ -- B: If $(x,y)$ is a solution, then so is $(x+b,y-a)$ – 2017-01-25

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