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Let $a,b$ be odd positive integers.

Suppose we have $$2\min(a,b)-\max(a,b)>0$$ $$2\min(a,b)-\max(a,b)=p$$ $$\min(a,b)\ne0\operatorname{mod}p$$ where $p$ is odd prime number.

Then I conjecture that $$\gcd(a,b)=1$$

Is there a way to prove it?

Notamathematician
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    Assume $a≤b$. So $2a-b=p$ If some prime $q$ divided $\gcd(a,b)$ then we'd have $q,|, p$, so.... – lulu Feb 17 '22 at 14:04
  • By $,a,b,$ symmetry, wlog $a\le b$ so the hypotheses become $[![1]!]\ \ldots, \ $ $[![2]!]\ \ 2a-b = p,\ $ $[![3]!]\ \ p\nmid a,,$ so $,(a,b) = (a,2a-p)\overset{\rm\color{#c00} R} = (a,-p) = 1,$ by $,p\nmid a,$ and $,\rm\color{#c00} R = $ gcd mod Reduction in the linked dupe. – Bill Dubuque Feb 17 '22 at 15:02

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WLOG, assume $b \geq a$. Then your second condition says that $2a = b + p$. Now suppose that $q$ is a prime that divides $a$. If it also divided $b$, then it would also have to divide $p$; but then we would have to have $q = p$. But your third condition says that $p \not\mid a$. Thus there is no prime that divides both $a$ and $b$, so $\gcd(a, b) = 1$.

Note that you do not need the first requirement, nor the requirement that $a, b$ are odd.

Mees de Vries
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