Can a coprime pair be decomposed into two coprime pairs?

Question

Suppose $2p>q>p>1$ and $(p, q)=1$ (i.e., $p$ and $q$ are coprime). Is it the case that we can always find $a,b, c,d$ such that

1. $p=a+c$, $q=b+d$,
2. $(a,b)=1$ and $(c,d)=1$,
3. $2a>b>a$ and $2c>d>c$

My attempt: first, some example: (5,7)=1 and $5=2+3$ and $7=3+4$. It holds.

A bit more general, if $q=p+2$, then $p$ must be even because $(p, q)=1$, so $p=2k+1$ and $q=2k+3$, it holds as well.

Some bigger (but random) example: $p=29, q=36$. Then $29=13+16$ and $36 =17+19$. It holds.

I guess one can work on $q=p+d$, but then I do not know how to proceed.

Answer 1

If $a<b<2a$ and $c<d<2c$, i.e., $a+1\le b\le 2a-1$ and $c+1\le d\le 2c-1$, then by adding these inequalities, we arrive at $a+c+2\le b+d\le 2a+2c-2$ or the necessary condition $$\tag1 p+2\le q\le 2p-2,$$ in other words, $q=p+1$ and $q=2p-1$ must be excluded.

In order to see that this restriction is also sufficient, note that $\frac qp$ is a fraction in shortest term and we have $1<\frac qp<2$. Consider the Farey sequence of all fractions in $[1,2]$ with denominator $\le p$. Then the obvious candidates for what you want are the immediate neighbours in this sequence: $$\frac11<\cdots<\frac ba<\frac qp<\frac dc <\cdots<\frac21$$ because then $\frac qp$ is the Farey sum $\frac{b+d}{a+c}$ (as is always the case for terms in a Farey sequence between terms with lower denominator). The only cases that do not work are when the we have left neighbour $\frac ba=\frac 11$ or right neighbour $\frac dc=\frac21$, i.e., when there is no fraction with denominator $\le p$ between $1$ and $\frac qp$ or between $\frac qp$ and $2$. But we certainly have $\frac{p+1}p\in ]1,\frac qp[$ and $\frac{2p-1}p\in]\frac qp,2[$ if $(1)$ holds.

In other words,

Given coprime integers $p,q$, there exist coprime integers $a,$ and coprime integers $c,d$ with $a<b< 2a$, $c<d<2c$, $p=a+c$, and $q=b+d$ if and only if $$p+2\le q\le 2p-2.$$