If gcd(a,b) then $ax+by=1$ also $cn=a+b$ for some $n$ because $c|a+b$.Therefore $cn-b=a$,substituting $cn-b=a$ in first equation we get $x(cn-b)+by=1=cnx-bx+by=c(nx)+b(y-x)$ which implies gcd(c,b)=1.Similarly gcd(c,a)=1.@Kartik Sharma how is it??Your's is good.

Note that $a|b$ implies that there exists an integer $k$ suck that $ka = b$, not $kb = a$.
So your assumption that $c = n(a+b)$ is wrong. It should be $cn = a+b$.

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## Comments

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TopNewest$c | (a+b) \rightarrow c|a \wedge c|b$ is wrong. Easily counterexample as $a = 4, b = 5, c = 3)$.

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Hint: use Bezout's identity

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Another hint: $(a,b) = 1$ if and only if $ax+by = 1$.

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@Samuraiwarm Tsunayoshi check my sol in Kartik Sharma's solution's replay box.

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Let $a = {{p}_{1}}^{{e}_{1}}{{p}_{2}}^{{e}_{2}}{{p}_{3}}^{{e}_{3}}..... {{p}_{n}}^{{e}_{n}}$

Now, $a = 0 mod {p}_{i}$

We know that gcd(a,b) = 1, so none of the ${p}_{i}$ divide b

$cn = a + b$

$cn mod {p}_{1} = 0 + x$

$cn = x mod {p}_{1}$

And so on, till ${p}_{n}$ which tells us that ${p}_{i}$ doesn't divide c. Therefore there is no common divisor of c and a, so, $gcd(a,c) = 1$

Similarly we can show for b.

@shivamani patil Check out and now I think it is fine.

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If

gcd(a,b)then $ax+by=1$ also $cn=a+b$ for some $n$ because $c|a+b$.Therefore $cn-b=a$,substituting $cn-b=a$ in first equation we get $x(cn-b)+by=1=cnx-bx+by=c(nx)+b(y-x)$ which impliesgcd(c,b)=1.Similarlygcd(c,a)=1.@Kartik Sharma how is it??Your's is good.Log in to reply

Yep, looks good.

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I knew about this method too but that is already discussed by Samuraiwarm Tsunayoshi.

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@Kartik Sharma

But I solved it independentlyLog in to reply

Note that $a|b$ implies that there exists an integer $k$ suck that $ka = b$, not $kb = a$. So your assumption that $c = n(a+b)$ is wrong. It should be $cn = a+b$.

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Oh damn yes ! I knew it but just a silly mistake. Now, I am gonna edit it.

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