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Ramanujan's Taxicab number 1729 is famous for being the smallest positive integer which can be written as the sum of two positive cubes in two different ways. On a different note, I observed that

$12^3 + 1^3 = 1729$

$12^2 + 1^2 = 1\cdot7\cdot2\cdot9 + 1 + 7 + 2 + 9$

$12^1 + 1^1 = -1+7-2+9$

Question 1: Is there any other $n$ number with the property that if $n = a^3 + b^3$ for some positive $a$ and $b$ then

$$ a^2 + b^2 = \text{Products of the digits of $n$} \ + \ \text{Sum of the digits of $n$} $$

Question 2: While searching for a solution of Question 1, the program run by Peter has found only two solution, $(6,11)$ and $(1,12)$ for $1 \le a,b \le 20000$. Looks like there are no more solution. Can this be proven or disproven?

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    $\begingroup$ The close-voter should , instead of voting for close, be glad that someone has the courage to post a puzzle here. I will never understand why such puzzles are not welcome here. (This question is an exception, $7$ upvotes are unusual). What I understand, that the forum is not a make-other-homeworks-forum, but what is bad about puzzles, I cannot imagine. $\endgroup$ Commented Oct 20, 2016 at 19:56
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    $\begingroup$ @Peter The close vote was cast before the edit specifying the property. The question was unclear then. I had commented and asked the OP to make the desired property clear. If the close voter didn't want to watch the question to see whether it would be clarified, the close vote was perfectly reasonable then. Now that the question is clear, further close votes wouldn't be appropriate. (Whether the close voter comes back and retracts the close vote, or the close vote is left to age away is unimportant.) $\endgroup$ Commented Oct 20, 2016 at 20:11
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    $\begingroup$ And I applaud you for clarifying the question in a timely fashion. $\endgroup$ Commented Oct 20, 2016 at 20:14
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    $\begingroup$ @DanielFischer OK, I didn' know this. I am sorry. But maybe this question changes the trend and more puzzles are posted here. $\endgroup$ Commented Oct 20, 2016 at 20:20
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    $\begingroup$ Lawyers think they are the champion pettifoggers, but mathematicians are the true masters. $\endgroup$ Commented Oct 20, 2016 at 20:33

3 Answers 3

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If I understood correctly the problem is

Find numbers $n$ such that there exist a pair $(a,b)$ with the property that

$a^3+b^3=n$

$a^2+b^2$=products of the digits of n + Sum of the digits of n

Then, the number $1547$ is a solution, for the pair $(a,b)=(11,6)$:

$11^3+6^3=1547\\11^2+6^2=1\cdot 5 \cdot 4 \cdot 7 + 1 + 5 + 4 + 7\\11^1+6^1 = 1+5+4+7$

And of course also the number $0$ is a solution. I also feel that there are no more solutions.

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    $\begingroup$ Seems, that you have been faster ... $\endgroup$ Commented Oct 20, 2016 at 20:00
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    $\begingroup$ @Pigrecoric was faster by only 19 seconds :) $\endgroup$ Commented Oct 20, 2016 at 20:02
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    $\begingroup$ No more solutions for $1\le a\le b\le 5000$ $\endgroup$ Commented Oct 20, 2016 at 20:02
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    $\begingroup$ No more solutions for $1\le a\le b\le 10000$ $\endgroup$ Commented Oct 20, 2016 at 20:08
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    $\begingroup$ $$1547 = 7\times 13\times 17$$ and $$1729 = 7 \times 13 \times 19$$ Both are simple products of 3 small primes. $\endgroup$ Commented Oct 20, 2016 at 20:46
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The first two conditions are satisfied by $$a=6\ \ \ \ , \ \ \ \ b=11$$

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  • $\begingroup$ The evil number $666$ comes close. $$\begin{aligned}666^3 + 542^3 &=n\\ 2^2+666^2 + 542^2 &=\text{digit product of n + digit sum of n}\end{aligned}$$ It took too long for my computer to go $a,b>1000$ to find large near-solutions. $\endgroup$ Commented Jan 16, 2018 at 18:16
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$(a,b)=(12,1)$ and $(a,b)=(11,6)$ are the only integer solutions with $2500 \geq a > b$.

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