A student recently asked me about solutions to the equation $$a_{b} = b_{a},$$ where the subscript notation $a_{b}$ denotes interpreting the digits of $a$ in base $b$. It turns out there are tons of non-trivial solutions to this equation, such as $$ 36_{49} = 49_{36},$$ which can be checked since $3(49) + 6 = 153 = 4*(36) + 9$. If you allow yourself "digits" larger than 9, you get even more examples. For one, using $C$ to represent 12 as in hexadecimal, $$ 18C_{270} = 270_{192}.$$ Here, that 192 is the number 18C interpreted in base 10 - that is, 100 + 80 + 12. Always interpreting the bases as if they themselves were written in base 10 is a convention, but a necessary one, I think.
Anyway, my question is whether anything is known about this equation. It seems simple to state, difficult to solve, and completely useless. :) Thus, I can't get it out of my head. If it's been looked at before by anyone, I'd love a reference or ideas about it.
Restricting myself to the 2 digit problem...
Require $pq_{xy}=xy_{pq}$ where we are working in base $d$.
$p \times (xy_d)+q=x \times (pq_d)+y$
$p \times (xd+y)+q=x \times (pd+q)+y$
$pdx+py+q=pdx+qx+y$
$py+q=qx+y$
$py-y=qx-x$
$y(p-1)=x(q-1)$
This explains why there are so many solutions: all you need is a number that can be factorised in at least two different ways.
Moving on to the 3 digit problem...
Require $pqr_{xyz}=xyz_{pqr}$ where we are working in base $d$.
$p \times (xyz_d)^2+q \times (xyz_d)+r=x \times (pqr_d)^2+y \times (pqr_d)+z$
$p (xd^2+yd+z)^2+q (xd^2+yd+z)+r=x (pd^2+qd+r)^2+y (pd^2+qd+r)+z$
$p (x^2d^4+y^2d^2+z^2+2xyd^3+2xzd^2+2yzd)+q (xd^2+yd+z)+r= x (p^2d^4+q^2d^2+r^2+2pqd^3+2prd^2+2qrd)+y (pd^2+qd+r)+z$
$px^2d^4+py^2d^2+pz^2+2pxyd^3+2pxzd^2+2pyzd+qxd^2+qyd+qz+r= xp^2d^4+xq^2d^2+xr^2+2xpqd^3+2xprd^2+2xqrd+ypd^2+yqd+yr+z$
... stopping now!