GCD property

Note by Shivamani Patil
4 years, 11 months ago

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

Samuraiwarm Tsunayoshi - 4 years, 11 months ago

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Hint: use Bezout's identity

Samuraiwarm Tsunayoshi - 4 years, 11 months ago

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

Samuraiwarm Tsunayoshi - 4 years, 11 months ago

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

shivamani patil - 4 years, 11 months ago

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Let a=p1e1p2e2p3e3.....pnena = {{p}_{1}}^{{e}_{1}}{{p}_{2}}^{{e}_{2}}{{p}_{3}}^{{e}_{3}}..... {{p}_{n}}^{{e}_{n}}

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

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

cn=a+bcn = a + b

cnmodp1=0+xcn mod {p}_{1} = 0 + x

cn=xmodp1cn = x mod {p}_{1}

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

Similarly we can show for b.

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

Kartik Sharma - 4 years, 11 months ago

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If gcd(a,b) then ax+by=1ax+by=1 also cn=a+bcn=a+b for some nn because ca+bc|a+b.Therefore cnb=acn-b=a,substituting cnb=acn-b=a in first equation we get x(cnb)+by=1=cnxbx+by=c(nx)+b(yx)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.

shivamani patil - 4 years, 11 months ago

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Yep, looks good.

Samuraiwarm Tsunayoshi - 4 years, 11 months ago

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

Kartik Sharma - 4 years, 11 months ago

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@Kartik Sharma But I solved it independently @Kartik Sharma

shivamani patil - 4 years, 11 months ago

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

Siddhartha Srivastava - 4 years, 11 months ago

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

Kartik Sharma - 4 years, 11 months ago

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@Kartik Sharma Looking forward to your solution!

Samuraiwarm Tsunayoshi - 4 years, 11 months ago

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