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Let $a,b,c$ are all integers greater than $1$.

Assume $\gcd(a,c)=1$, $\gcd(b,c)=1$ and $\gcd(b, c)=1$.

I am looking for $s$ which makes $\gcd(a+b, sc)=1$?

In general , $\gcd(a,c)=1$, $\gcd(b,c)=1$ doesn't imply $\gcd(a+b,c)=1$. For example, take $a=3, b=4, c=7$, then $\gcd(a,c)1,~\gcd(b,c)=1$ but $a+b=7=c$.

Therefore, $\gcd(a+b,sc)=1$ is also not true in general, for any integer $c$.

However, if I choose especial forms like $a=\frac{p^k+1}{2}, ~b=\frac{p^l+1}{2}, ~c=\frac{p^m-1}{2}$ and $m$ is odd and $p$ is prime number, $k,l$ are positive integers, then it can be checked that $\gcd(a,c)=1=\gcd(b,c)$. In that case if $s=p^m$, I have checked for $(k,l,m)=(2,3,5)$ or $(3,4,7)$ $$\gcd(a+b,sc)=1.$$

I am looking for all such choices of $a,b,c,s$ such that $\gcd(a+b,sc)=1$.

Comment: This is some edit of the previous version of the question, where I didn't mention such examples and that is why some people in the comment section downvoted this as invalid question. But I believe it is eligible to be fitted here.

Martin Sleziak
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1 Answers1

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As far as I understand, there are two ways to interpret the objective in this problem:

  • finding constant integers $r,s$ such that $\gcd(ra+b,sc) = 1$ for all pairwise coprime positive integers $a,b,c$, or
  • given pairwise coprime integers $a,b,c$, finding all integers $r,s$ that satisfies $\gcd(ra+b,sc) = 1$ in terms of $a,b$ and $c.$

For the first interpretation, it can be easily proved that no pair of integers $(r,s)$ satisfy that condition. If there is a prime $p$ that divides both $r$ and $s$, take $b=p$ and take $a$ and $c$ to be any two coprime integers that are not multiples of $p$. Otherwise, take $b=1$. Bezout's theorem ensures that there are coprime integers $a,c$ such that $ra+1=sc$.

For the second interpretation, I'm not aware of a method that categorizes all such pairs $(r,s)$. But we can generate a fair amount of them. It is sufficient to choose $r$ and $s$ such that each prime divisor of $sc$ divides exactly one of the integers $ra$ or $b$. Since no prime divisor of $c$ can divide $a$ or $b$, taking $r = p_1^{\alpha_1} p_2^{\alpha_2} \cdots$ and $s = q_1^{\beta_1} q_2^{\beta_2} \cdots$ works where $\alpha_i$ and $\beta_i$ are any positive integers, $p_i$ are all prime divisors of $c$, and $q_i$ are some prime divisors of $b$.

Fikilis
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