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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