Prove that if gcdb c 1 then gcda bc gcda bgcda cSolutionUs

Prove that if gcd(b, c) = 1, then gcd(a, bc) = gcd(a, b)gcd(a, c)

Using Bezout\'s Identity:

Since there are x,y,u,v so that ax+by=1 and au+cv=1 , we have

bycva(x+u?axu)+bcvy =(1?ax)(1?au)=1?a(x+u?axu)=1

Therefore, (a,bc)=1

Since a,b are relatively prime, there exist integers x,y such that ax+by=1 . Multiplying both sides of this equation by c , we get

acx+bcy=c.

Using Bezout\'s Identity:

Since there are x,y,u,v so that ax+by=1 and au+cv=1 , we have

bycva(x+u?axu)+bcvy =(1?ax)(1?au)=1?a(x+u?axu)=1

Therefore, (a,bc)=1

Since a,b are relatively prime, there exist integers x,y such that ax+by=1 . Multiplying both sides of this equation by c , we get

acx+bcy=c.

Letting d=gcd(a,bc) , we see from the equation above that since d is a common divisor of a and bc , d also divides c by linearity. Then, since every common divisor of two integers also divides their gcd, d|gcd(a,c) . But gcd(a,c)=1 , so d divides 1 . Because the gcd is nonnegative, it follows that d=gcd(a,bc)=1 .

$Prove that if gcd(b, c) = 1, then gcd(a, bc) = gcd(a, b)gcd(a, c)SolutionUsing Bezout\'s Identity: Since there are x,y,u,v so that ax+by=1 and au+cv=1 , we have$

Prove that if gcdb c 1 then gcda bc gcda bgcda cSolutionUs

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