If ax + by = 1,where a, b, x and y are intege | vedclass

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Question

(C) The equation $ax + by = 1$ is a linear Diophantine equation.According to Bezout's identity,the equation $ax + by = c$ has integer solutions for $x

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$(b, y) = 1$

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(C) The equation $ax + by = 1$ is a linear Diophantine equation.According to Bezout's identity,the equation $ax + by = c$ has integer solutions for $x$ and $y$ if and only if $\gcd(a, b)$ divides $c$.Here,$c = 1$,so $\gcd(a, b)$ must divide $1$,which implies $\gcd(a, b) = 1$. Thus,option $(d)$ is true.Similarly,for $ax + by = 1$,we can rewrite it as $ax = 1 - by$,which implies $\gcd(a, b) = 1$ and $\gcd(a, y) = 1$ is not necessarily true,but $\gcd(a, b) = 1$ is a necessary condition.Specifically,$\gcd(a, b) = 1$,$\gcd(a, x)$ is not necessarily $1$,$\gcd(b, x)$ is not necessarily $1$,and $\gcd(x, y)$ is not necessarily $1$.However,looking at the options,$\gcd(b, y) = 1$ is not a required condition for the existence of solutions to $ax + by = 1$. For example,if $a=1, b=2, x=1, y=0$,then $1(1) + 2(0) = 1$. Here $\gcd(b, y) = \gcd(2, 0) = 2 
eq 1$. Thus,option $(c)$ is not necessarily true.

If $ax + by = 1$,where $a, b, x$ and $y$ are integers,then which one of the following is not true?

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