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Dear Mr Oliver,

I was wondering if you could help solve an argument. The equation is x^3 - y^3 = 5803. I got an answer of x = 20, y = 13 and x = -13, y = -20. My friend says to solve this, just set y^3 = 0 and then the answer is y = 0, x = cube root 5803. There seems to be something wrong with his answer, but I can't show mathematically if he's right or wrong and why. Would you be able to help clarify things?

Sincerely,

Alex D

It is hard to respond without knowing where this equation comes from.

x^3 - y^3 = 5803

That equation is a relation between x and y. I say "relation" and not "function" because depending on which number system you use, you may not be able to solve uniquely for y=f(x).

In complex numbers, for instance, for any y value there will be up to three corresponding x values so that x^3-y^3=5803.

Now, your friend is clearly using the real number system, over which picking y uniquely determines x. And that is his approach -- just, you know, pick any value of x and then solve for y by taking a cube root. He has just chosen x=0 as a convenience so that he doesn't have to do any subtraction.

Now, technically that is correct, his answer of ( 0 , 5803^(1/3) ) fits the equation. If you allow complex numbers, there are two more, namely ( 0 , 5803^(1/3) e^(2 π i /3) ) and another that is also, similarly, complex.

But those answers are only technically correct. The relation in question is not especially interesting or meaningful as a piece of algebra. What this relation probably is meant to describe is a

For example, the very famous Fermat's Last Theorem is a Diophantine equation. It states that there are no positive integer solutions to x^n + y^n = z^n for any fixed value of n. However, it is obvious and meaningless to observe that if x=0 and y=1, then z=1. Or if x=1 and y=1, then z = 2^(1/n). These are not positive integers, and they are not relevant to the problem.

I strongly suspect the intent of x^3-y^3=5803 is to be a Diophantine equation. As I said, if this question is literally "Find any two real numbers that solve this equation" then your friend is right. That could be it. I don't know where the problem came from. But I am going to assume he is wrong, and that there is more too it -- that it is a Diophantine equation.

For your equation, negative and zero values of x and y are feasible, and you may have noticed that if (a,b) is a solution, so is (-b,-a). You've already got at least one good solution.

One thing that is helpful is to approach this problem graphically. Notice in the attached image, there is a graph of the

As it turns out, they are the only two integer points on this curve, and here's why:

First of all, you can see from the drawing that if x and y are very large, they are very close to the line y=x (which is the dashed line). Indeed, if x is very large, y is also very large, one would say that "asymptotically" x and y are the same. (This is the same as saying there is an asymptote of this curve along the line y=x, which is also true.) So eventually, there is no way they could both be integers, because they are too close the a line that's chocked full of integer points.

More precisely, x^3-y^3 is a difference of two cubes (cubes of integers). These cubes get further and further apart as time goes on. To be precise, at some point, if n is large and you want x=n to be a part of a solution, then y=n-1 might not even be close enough. What if x=n and y=n-1 but you have:

x^3 - y^3 > 5803 ?

That can happen if n is too large. Why? Because:

x^3 - y^3 = n^3 - (n-1)^3 = 3 n^2 - 3n + 1

You can use the quadratic formula to conclude that 3n^2 - 3n + 1 > 5803 for n>44 (or n<-43).

Either way, if there are any integer points, then x and y are between -44 and 44. Now, that's still 89^2 = 7921 possible points, but a computer can check them all in only a few seconds. As it turns out, the two points you found are indeed the

So, in this way, you are indeed right and your friend -- while technically correct -- is looking at this problem from a wrong (or at least, mathematically uninteresting) perspective.

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