Help with Solving Diophantine Equations: Is Brute-Force the Only Option?

[jobsism]

EDIT:- My question earlier was in error. I've edited the question accordingly and my required set of Diophantine equations is the one given below:-

For a 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 \notin {2, 9, 15, 16, 33, 34}.

How does one go about solving this? Is brute-force the only possible way? Or could there be a case that no integer solutions exist for this equation?

Also, are there any online computing engines, that allow me to set constraints, and solve Diophantine equations of this sort?

Any and all help is appreciated! Thanks!

 

[tiny-tim]

hi jobsism!

let's go for the easy equation …

a³ - b³ = 721 …

doesn't that make it obvious what a and b must be? ​

 

[Ryuzaki]

a³ - b³ = 721 …

doesn't that make it obvious what a and b must be? ​

Don't you mean b^3 - a^3 = 721, tiny-tim?

Your method is standard, but it requires checking and cross-checking for quite a number of values.

To the OP: In case you didn't get tiny-tim's hint, what he's trying to say is that b^3 - a^3 = 721 gives (b-a)(a^2 + ab + b^2) = 7x103, and so (b-a) \in {1,7,103,721}. Then you compute a and b for each of these values, get the corresponding d values, and plug it in the equation, c^3 +4104 = d. This is a rather tedious method in my opinion, but it would work.

However, I suggest a slightly easier approach. First, notice that for any solution, b^3 - a^3≥(a+1)^3 - a^3 = 3a^2 + 3a + 1. This gives a direct upper bound on a, namely 15. Now, checking for which of a = 1,2,3,...15, one has that a^3 + 721 is the third power of an integer, one finds that this is only the case for a=2 and a=15 (where b would be 9 and 16 respectively).

Both the above solutions are rejected, as per your constraints. So already, your first two equations do not have any solutions.

 

[tiny-tim]

Hi Ryuzaki!

To the OP: In case you didn't get tiny-tim's hint, what he's trying to say is that b^3 - a^3 = 721 gives (b-a)(a^2 + ab + b^2) = 7x103, and so (b-a) \in {1,7,103,721}. Then you compute a and b for each of these values, get the corresponding d values, and plug it in the equation, c^3 +4104 = d. This is a rather tedious method in my opinion, but it would work.

Slightly easier would be eg if b - a = 1, then (b - a)² - 3ab = 721, so 3ab = 720, ab = 240, (a,b) = (15,16), which the question doesn't allow