Help with this system of Diophantine equations

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!

• 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? May 21, 2013 at 18:20

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.

• I redecided and fixed, it is hoped, the answer, not to create additional complaints regarding me embarrassing the site by my sloppiness ;-)
– user9072
May 21, 2013 at 16:58
• Note, if you don't worry about the exclusions, the corresponding values for $c$ are $33$ and $34$ (which are also excluded). May 21, 2013 at 17:01
• @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).
– user9072
May 21, 2013 at 17:10
• 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! :) May 21, 2013 at 17:10
• @Jobin Idiculla: You are welcome!
– user9072
May 21, 2013 at 17:11

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.

• quid found a slicker approach (although it too requires going through a bunch of cases). May 21, 2013 at 17:03
• Thanks, but in my opinion yours is more elegant.
– user9072
May 21, 2013 at 17:06
• Thank you, Barry. Now I understand the general approach to solving such problems. May 21, 2013 at 17:09
• 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. May 21, 2013 at 20:57
• @Joe: you made me curious about the smallest $|m|$ for which equation $a^3 - 2\cdot b^3 = m$ is already hard. May 22, 2013 at 6:58