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Note that we have:

$$A=\{(a,b)\in\mathbb{Q}\times\mathbb{Q}~|~\text{Both equations}~a+x=b, b+y=a~\text{have answers in }~\mathbb{Q}\}=\mathbb{Q}\times\mathbb{Q}$$

$$B=\{(a,b)\in\mathbb{Q}\times\mathbb{Q}~|~\text{Both equations}~a\times x=b, b\times y=a~\text{have answers in}~\mathbb{Q}\}=(\mathbb{Q}\setminus \{0\})\times (\mathbb{Q}\setminus \{0\})\cup \{(0,0)\}$$

Now consider:

$$C=\{(a,b)\in\mathbb{Q}\times\mathbb{Q}~|~\text{All equations}~a^x=b, b^y=a, z^a=b, t^b=a~\text{have answers in}~\mathbb{Q}\}$$

$C$ is non-empty trivially because $(1,1)\in C$, is $C$ infinite?

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    Rational powers of negative rational numbers aren't even (always) defined; your question is not clear about how to interpret this in the definition of $C$. – Marc van Leeuwen Mar 19 '15 at 09:50
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    Note that, if $a^x=b$, we also have $b^{(1/x)}=a$, $(a^{x/a})^a=b$, and $(a^{1/b})^b=a$. You want these all to be rational, I guess? – Akiva Weinberger Mar 19 '15 at 09:50
  • @columbus8myhw Yes, everything is rational. –  Mar 19 '15 at 11:38
  • @MarcvanLeeuwen Clearly I mean where the equations are meaningful and have answers. Note that your question is like the case of the set $B$ when $0$ comes to the story. –  Mar 19 '15 at 12:59
  • It might be useful to look at the larger set $D={(a,b)\in\mathbb Q\times\mathbb Q\ |\ (a^{1/b}\in\mathbb Q)\land(b^{1/a}\in\mathbb Q)}$, since $C$ is a subset of $D$, and $D$ looks easier to analyze. (And then we can use what we find to understand $C$ better.) – Akiva Weinberger Mar 19 '15 at 16:23
  • In addition to @MarcvanLeeuwen's answer, we have stuff like $(a,b)=(\frac13,\frac19)$, since we can have $(x,y,z,t)=(2,\frac12,\frac1{9^3},\frac1{3^9})$. (In case you're curious, $9^3$ and $3^9$ are $729$ and $19683$ respectively.) – Akiva Weinberger Mar 20 '15 at 20:39

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$C$ is indeed infinite: it contains all pairs $(a,b)=(\frac1n,\frac1n)$ for integer $n>0$; solutions are $x=y=1$ and $z=t=n^{-n}$.

Marc van Leeuwen
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