Adopting the following notation:

$$ R(b/a) = \text{Remainder of b when divided by a} $$

So I was trying to prove the following:

There always exists an $n$ for two primes $a$ and $b$ such that:

$$ R(b/a) = R(2n/a) $$

and satisfies:

$$ 2n > b > n > a > 2$$

  • 1
    $\begingroup$ Sure. What about $2n=a+b$? $\endgroup$ – Jef L Nov 11 '15 at 9:37
  • $\begingroup$ You'd get further if you just wrote the question asking for $n$ such that $b\equiv 2n\pmod {a}$. The remainder notation is not clarifying, and $\equiv$ notation is quite standard. As Jef said, if you have $a,b$ odd primes, then $a+b$ is even and $n=(a+b)/2$ satisfies your condition. Of course, you didn't even assume $b>a$, so it is not always possible for find $b>n>a$ unless you add conditions. $\endgroup$ – Thomas Andrews Nov 17 '15 at 16:52

I think the proposition is valid for any two odd numbers:

Let $b=a+2q$ and $n=a+q$ we have $2n=2a+2q$. So $R(b,a)=2q=R(2n,a)$


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