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Intuitively I would say yes but I can't find a way to prove it. I tried with small values and bruteforcing shows that there seems to only be one solution given a distinct tuplet.

For example $(1,72)$ has only $(8,9)$ as valid $a$ and $b$ values. Is there a way to do this mathematically?

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    $\begingroup$ Welcome to Puzzling.SE! For the uninitiated in the room (such as me), would you mind explaining what precisely is meant by the little symbol that looks like a Phillips head screw? Thanks, and have a great day! $\endgroup$ – Brandon_J May 4 '19 at 17:28
  • $\begingroup$ It’s bitwise addition modulo 2, @Brandon_J $\endgroup$ – El-Guest May 4 '19 at 17:30
  • $\begingroup$ Thanks for letting me know, @El-Guest . $\endgroup$ – Brandon_J May 4 '19 at 17:30
  • $\begingroup$ So essentially, I convert the number to binary (for exapmle, 3-->00011, 24-->11000), then perform theXOR operation on each pair of bits, and then convert back to a normal number (27)? $\endgroup$ – Brandon_J May 4 '19 at 17:54
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    $\begingroup$ Not sure this is a puzzle, but I'm glad you got your answer. (Please don't forget to $\color{green}{\checkmark \small\text{Accept}}$ it!) $\endgroup$ – Rubio♦ May 5 '19 at 4:34
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By counterexample, $a,b$ pair is clearly not unique.

The pairs $(5,9)$ and $(3,15)$ both multiply to $45$, and add bitwise to $12$.

$5 \oplus 9 = 12$, $5 \times 9 = 45$
$3 \oplus 15 = 12$, $3 \times 15 = 45$

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    $\begingroup$ @Hugh, thanks for the edits - looks nicer now! $\endgroup$ – ppgdev May 5 '19 at 3:09
  • $\begingroup$ I edited it quickly off my phone a few hours ago, so I have it another go. Hopefully it still looks good. $\endgroup$ – anon_user May 5 '19 at 4:47
  • $\begingroup$ @Hugh, sure. Thanks. $\endgroup$ – ppgdev May 5 '19 at 16:28

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