Further investigation of the Mordell Equation

This post carries on from the previous post on the Mordell Equation – so make sure you read that one first – otherwise this may not make much sense.  The man pictured above (cite: Wikipedia) is Louis Mordell who studied the equations we are looking at today (and which now bear his name).

In the previous post I looked at solutions to the difference between a cube and a square giving an answer of 2.  This time I’ll try to generalise to the difference between a cube and a square giving an answer of k.  I’ll start with the same method as from the previous post:

In the last 2 lines we outline the 2 possibilities, either b = 1 or b = -1.  First let’s see what happens when b = 1:

This will only provide an integer solution for a if we have:

Which generates the following first few values for k when we run through m = 1, 2,3..:

k = 2, 11, 26, 47

We follow the same method for b = -1 and get the following:

Which generates the following first few values for k when we run through m = 1, 2,3…:

k = 4, 13, 28, 49

These are the values of k which we will be able to generate solutions to. Following the same method as in the previous post this generates the following solutions:

Let’s illustrate one of these results graphically.  If we take the solutions for k = 13, which are (17,70) and (17,-70), these points should be on the curve x cubed – y squared = 13.

This is indeed the case.  This graph also demonstrates how all solutions to these curves will have symmetrical solutions (e, f) and (e, -f).

We can run a quick computer program to show that this method does not find all the solutions for the given values of k, but it does ensure solutions will be found for the k values in these lists.

In the code solutions above, results are listed k, x, y, x cubed, y squared.  We can see for example that in the case of k = 11 our method did not find the solution x = 3 and y = 4 (though we found x = 15 and y = 58).  So, using this method we now have a way of finding some solutions for some values of k – we’ve not cracked the general case, but we have at least made a start!