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I proceeded this way:

gcd(a,b) = 1 leads to ax + by = 1

gcd(b,c) = 1 leads to cl + dm = 1

a/b = c/d = k and ax+by = cl+dm

Then I trying to operate the equation so that I end up finding k=1 but Im stuck at:

lets a = ck and b = dk

ckx + dky = cl + dm

k(cx+dy) = cl + dm

k = 1/(cx+dy)

and nothing tells me that cx+dy=1 since we have to prove that a=c and b=d

Help please?

Steven
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    You are probably overthinking. First I think $a,b,c,d$ are positive integers (otherwise the claim is false). Just write $ad=bc$. Use the fact that $\gcd(a,b)=1$ to show that $a\mid c$. Use the fact that $\gcd(c,d)=1$ to show that $c\mid a$. – Batominovski Jan 29 '20 at 20:50

1 Answers1

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Hint

$$ad=bc\implies {a|c\to c=ak_1\\ d|b\to b=dk_2}$$Now, since $ad=bc$, what do we conclude?

Mostafa Ayaz
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