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A couple hours ago, I'd posted a Diophantine equation question, but realized that I'd committed a rather preposterous blunder deriving it.

This is the actual question which I'm trying to solve:-

For a research problem that I'm working on, I need to solve the following system of Diophantine equations:-

$ a^3 + 40033 = d$, $ b^3 + 39312 = d$, $ c^3 + 4104 = d$, where $a,b,c,d>0$ are all DISTINCT positive integers and $a,b,c∉${$2,9,15,16,33,34$}.

Sorry about the earlier mishap. Any and all help is appreciated! Thanks!

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  • $\begingroup$ Where did these excluded values come from? Do you know why they consist exactly of all the values that occur in the solutions prior to exclusion? $\endgroup$
    – Will Sawin
    May 21, 2013 at 18:20

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The first two equations amount to $b^3 - a^3 = 721$.

Now since for any solution $b^3 - a^3 \ge (a+1)^3 - a^3 = 3a^2 + 3a +1$ this directly gives an upper bound on $a$ namely $15$.

Now checking for which of $a=1, \dots, 15$ one has that $a^3 + 721$ is the third power of an integer, by calculating its third root for example one finds that this is only the case for $a=2$ and $a=15$ (where $b$ would be $9$ and $16$, respectively).

Both are excluded, so already the first two equations have no solution.

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  • $\begingroup$ I redecided and fixed, it is hoped, the answer, not to create additional complaints regarding me embarrassing the site by my sloppiness ;-) $\endgroup$
    – user9072
    May 21, 2013 at 16:58
  • $\begingroup$ Note, if you don't worry about the exclusions, the corresponding values for $c$ are $33$ and $34$ (which are also excluded). $\endgroup$ May 21, 2013 at 17:01
  • $\begingroup$ @Barry Cipra: thank you for this additional information (which I did not check myself). So without the restriction the full set of solutions is, giving (a,b,c): (2,9,33) and (15,16,34). $\endgroup$
    – user9072
    May 21, 2013 at 17:10
  • $\begingroup$ Thanks a lot, quid! Your solution is more elegant (and it also saved me a lot of time!) Thanks, you guys! Learned a lot today! :) $\endgroup$ May 21, 2013 at 17:10
  • $\begingroup$ @Jobin Idiculla: You are welcome! $\endgroup$
    – user9072
    May 21, 2013 at 17:11
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I'm not going to do all the calculations, just get things started.

From $a^3+40033 = d = b^3+39312$, one gets $b^3-a^3=721$, or $(b-a)(b^2+ab+a^2)=7\cdot103$, so $b-a\in 1,7,103,721$. You can do these cases one at a time. For example, if $b-a=1$ then you have $(a+1)^2+a(a+1)+a^2=721$, or $a^2+a-240=0$, which has no solutions. If one of the cases does produce a solution, then you can compute the corresponding $d$ and check it against the equation $d=c^3+4104$.

There may well be a slicker approach, but this should work.

Added 5/22/13: I just discovered a mildly embarrassing error in my answer. The equation $a^2+a-240=0$ does have solutions: The quadratic factors as $(a-15)(a+16)$. (I had mentally multiplied $4\times240=920$ and knew that a discriminant of $921$ was too close to $900$ to be a perfect square.) I finally caught my error when I decided to try to make my approach a little slicker:

If you write $b=a+k$, then $b^3-a^3=721$ becomes $k(k^2+3ka+3a^2)=7\cdot103$, so again $k \in 1,7,103,721$. But now you can immediately rule out $k=103$ and $k=721$, because the quadratic factor $k^2+3ka+a^2$ would obviously produce a number way too large (since $a$ is also required to be positive). So that leaves $k=1$ and $k=7$, both of which do lead to solutions. It so happens, though, that they lead to $(a,b,c)=(15,16,34)$ and $(2,9,33)$, respectively, which the OP explicitly disallowed.

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  • $\begingroup$ quid found a slicker approach (although it too requires going through a bunch of cases). $\endgroup$ May 21, 2013 at 17:03
  • $\begingroup$ Thanks, but in my opinion yours is more elegant. $\endgroup$
    – user9072
    May 21, 2013 at 17:06
  • $\begingroup$ Thank you, Barry. Now I understand the general approach to solving such problems. $\endgroup$ May 21, 2013 at 17:09
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    $\begingroup$ It's always intrigued me that for a given $m$, solving $a^3-b^3=m$ is easy (at least if we can factor $m$), but solving $a^3-2b^3=m$ is very difficult. An "intrinsic" explanation is that $\mathbb Z$ has only two units, while $\mathbb Z[2^{1/3}]$ has infinitely many units. But still, its amazing that there was no general effective solution method for $a^3-2b^3=m$ until Baker's theorem. $\endgroup$ May 21, 2013 at 20:57
  • $\begingroup$ @Joe: you made me curious about the smallest $|m|$ for which equation $a^3 - 2\cdot b^3 = m$ is already hard. $\endgroup$ May 22, 2013 at 6:58

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