# Remainder of prime = Remainder of an even number?

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

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:

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

• Sure. What about $2n=a+b$? – Jef L Nov 11 '15 at 9:37
• 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. – Thomas Andrews Nov 17 '15 at 16:52

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