if $a$ is prime to $b$ and $y$, $b$ is prime to $x$ then prove that $ax+by$ is prime to $ab$

Question

I think this question is incorrect as:

let $ax+by = p, then, (a,b)|p, (a,y)|p, (x,b)|p, (x,y)|p$

now : $(a,b) = (a,y) = (x,b) = 1$

now let $p = m(x,y)$

so left to prove $(ax + by, ab) = 1$

which is the same as: $(m(x,y),ab) = 1$

but $m$ might be equal to any factor or multiple of $ab$

so I don't understand how is this true

(I'm sorry if my question tag is wrong as I couldn't figure out what to tag it with.)

Does the loose language "is prime to" mean "is coprime with?" Do you mean $a$ is coprime with $b$ and $y$ and $b$ is coprime with $x$, and the statement is prove that $ax+by$ is coprime with $ab$? — FShrike, Jul 31 '21 at 18:15
it did solve it but i dont see any flaws in my approach but its still not working — AYUSHROCKS123, Jul 31 '21 at 18:22

score 1 · Answer 1 · answered Jul 31 '21 at 18:53

1

Since $(a,b)=1$ we have

$(ax+by, ab) = (ax+by,a)(ax+by,b)$

But $(ax+by,a) = (by,a)=1$ and $(ax+by,b)=(ax,b)=1$

answered Jul 31 '21 at 18:53

jjagmath

18,214

your explanation is correct and I got the proof but if you consider my method I'm going wrong nowhere but still am able to prove that it is not true so I want an explanation on that – AYUSHROCKS123 Aug 01 '21 at 07:28
Apparently, you are under the impression that you may choose $k$ to be any integer you want. That's not the case. – jjagmath Aug 01 '21 at 10:18
do you mean m?? – AYUSHROCKS123 Aug 01 '21 at 12:18
Yes, i meant $m$, sorry – jjagmath Aug 01 '21 at 12:39
then what restrictions can there exist upon choosing of m I thought about it but didn't get anything could you help me – AYUSHROCKS123 Aug 02 '21 at 16:45

score 0 · Answer 2 · answered Jul 31 '21 at 18:55

Since $a$ is coprime to both $b$ and $y$, it is also coprime to $by$, and hence to $ax+by$ since adding multiples of $a$ does not change coprimality. Likewise, since $b$ is coprime to both $a$ and $x$, it is also coprime to $ax$, and hence to $ax+by$ since adding multiples of $b$ does not change coprimality. Hence, $ab$ is coprime to $ax+by$ since it is the product of two numbers coprime to $ax+by$.

This proof only requires using two facts:

If $m$ and $n$ are coprime to $k$, then so is $mn$.
If $m \equiv n \pmod k$, then $m$ is coprime to $k$ if and only if $n$ is.

The first fact is used three times (first with $(m,n,k)=(b,y,a)$, then with $(m,n,k)=(a,x,b)$, and finally with $(m,n,k)=(a,b,ax+by)$).

The second fact is used twice (first with $(m,n,k)=(ax+by,by,a)$ and then with $(m,n,k)=(ax+by,ax,b)$).

your explanation is correct and I got the proof but if you consider my method I'm going wrong nowhere but still am able to prove that it is not true so I want an explanation on that — AYUSHROCKS123, Aug 01 '21 at 07:28

if $a$ is prime to $b$ and $y$, $b$ is prime to $x$ then prove that $ax+by$ is prime to $ab$

2 Answers2