Professor Tao,

I do not know whether this answer is in the least useful but will post it anyway!

I'm not sure but perhaps one approach is via Linnik's theorem that the least prime, say, $p(r,q)$ in an arithmetic progression $r \bmod q$, is $\ll q^L$.

(I actually saw that topic on Math Overflow recently: least prime in a arithmetic progression )

If $p \equiv -b \bmod a$ and $p \equiv -1 \bmod b$, with $(a,b)=1$ and $b$ of size about $a^2$, then one can take

$$ m = \frac{p+b}{a} $$

and

$$ \overline{m} = \frac{a(p+1)}{b}. $$

But then $p$ is in an arithmetic progression of modulus of size about $a^3$, so

$$ \max(m,\overline{m}) \ll p^{1-1/(3L)}. $$

However, currently the state of knowledge of $L$ is not good enough.

EDIT:

Actually, one can take $a,b$ about the same size, $(a,b)=1$, $p\equiv -b \bmod a$ and $p \equiv -a \bmod b$, and

$$ m = \frac{p+b}{a} $$

and

$$ \overline{m} = \frac{p+a}{b}. $$

Then $p$ is in an arithmetic progression of modulus of size about $a^2$, and

$$ \max(m,\overline{m}) \ll p^{1-1/(2L)}. $$

Professor Tao,

I do not know whether this answer is in the least useful but will post it anyway!

I'm not sure but perhaps one approach is via Linnik's theorem that the least prime, say, $p(r,q)$ in an arithmetic progression $r \bmod q$, is $\ll q^L$.

(I actually saw that topic on Math Overflow recently: least prime in a arithmetic progression )

If $p \equiv -b \bmod a$ and $p \equiv -1 \bmod b$, with $(a,b)=1$ and $b$ of size about $a^2$, then one can take

$$ m = \frac{p+b}{a} $$

and

$$ \overline{m} = \frac{a(p+1)}{b}. $$

But then $p$ is in an arithmetic progression of modulus of size about $a^3$, so

$$ \max(m,\overline{m}) \ll p^{1-1/(3L)}. $$

However, currently the state of knowledge of $L$ is not good enough.

EDIT:

Actually, one can take $a,b$ about the same size, $(a,b)=1$, $p\equiv -b \bmod a$ and $p \equiv -a \bmod b$, and

$$ m = \frac{p+b}{a} $$

and

$$ \overline{m} = \frac{p+a}{b}. $$

Then $p$ is in an arithmetic progression of modulus of size about $a^2$, and

$$ \max(m,\overline{m}) \ll p^{1-1/(2L)}. $$

Professor Tao,

I do not know whether this answer is in the least useful but will post it anyway!

I'm not sure but perhaps one approach is via Linnik's theorem that the least prime, say, $p(a,q)$ in an arithmetic progression $a \bmod q$, is $\ll q^L$.

(I actually saw that topic on Math Overflow recently: least prime in a arithmetic progression )

If $p \equiv -b \bmod a$ and $p \equiv -1 \bmod b$, with $(a,b)=1$ and $b$ of size about $a^2$, then one can take

$$ m = \frac{p+b}{a} $$

and

$$ \overline{m} = \frac{a(p+1)}{b}. $$

But then $p$ is in an arithmetic progression of modulus of size about $a^3$, so

$$ \max(m,\overline{m}) \ll p^{1-1/(3L)}. $$

However, currently the state of knowledge of $L$ is not good enough.

Professor Tao,

I do not know whether this answer is in the least useful but will post it anyway!

I'm not sure but perhaps one approach is via Linnik's theorem that the least prime, say, $p(r,q)$ in an arithmetic progression $r \bmod q$, is $\ll q^L$.

(I actually saw that topic on Math Overflow recently: least prime in a arithmetic progression )

If $p \equiv -b \bmod a$ and $p \equiv -1 \bmod b$, with $(a,b)=1$ and $b$ of size about $a^2$, then one can take

$$ m = \frac{p+b}{a} $$

and

$$ \overline{m} = \frac{a(p+1)}{b}. $$

But then $p$ is in an arithmetic progression of modulus of size about $a^3$, so

$$ \max(m,\overline{m}) \ll p^{1-1/(3L)}. $$

However, currently the state of knowledge of $L$ is not good enough.

EDIT:

Actually, one can take $a,b$ about the same size, $(a,b)=1$, $p\equiv -b \bmod a$ and $p \equiv -a \bmod b$, and

$$ m = \frac{p+b}{a} $$

and

$$ \overline{m} = \frac{p+a}{b}. $$

Then $p$ is in an arithmetic progression of modulus of size about $a^2$, and

$$ \max(m,\overline{m}) \ll p^{1-1/(2L)}. $$