@Kartik Sharma
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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 implies gcd(c,b)=1.Similarly gcd(c,a)=1.@Kartik Sharma how is it??Your's is good.
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Shivamani Patil
·
2 years, 3 months ago

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@Shivamani Patil
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I knew about this method too but that is already discussed by Samuraiwarm Tsunayoshi.
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Kartik Sharma
·
2 years, 3 months ago

@Kartik Sharma
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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 \( cn = a+b \).
–
Siddhartha Srivastava
·
2 years, 3 months ago

## Comments

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TopNewest\(c | (a+b) \rightarrow c|a \wedge c|b\) is wrong. Easily counterexample as \(a = 4, b = 5, c = 3)\). – Samuraiwarm Tsunayoshi · 2 years, 3 months ago

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Bezout's identity – Samuraiwarm Tsunayoshi · 2 years, 3 months ago

Hint: useLog in to reply

– Samuraiwarm Tsunayoshi · 2 years, 3 months ago

Another hint: \((a,b) = 1\) if and only if \(ax+by = 1\).Log in to reply

@Samuraiwarm Tsunayoshi check my sol in Kartik Sharma's solution's replay box. – Shivamani Patil · 2 years, 3 months ago

Hey,Log in to reply

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. – Kartik Sharma · 2 years, 3 months ago

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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. – Shivamani Patil · 2 years, 3 months agoLog in to reply

– Kartik Sharma · 2 years, 3 months ago

I knew about this method too but that is already discussed by Samuraiwarm Tsunayoshi.Log in to reply

@Kartik Sharma – Shivamani Patil · 2 years, 3 months ago

But I solved it independentlyLog in to reply

– Samuraiwarm Tsunayoshi · 2 years, 3 months ago

Yep, looks good.Log in to reply

– Siddhartha Srivastava · 2 years, 3 months ago

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 \( cn = a+b \).Log in to reply

– Kartik Sharma · 2 years, 3 months ago

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

– Samuraiwarm Tsunayoshi · 2 years, 3 months ago

Looking forward to your solution!Log in to reply