The Big Programming Thread - Page 715

On April 02 2016 00:34 Manit0u wrote:
Anyone here good with maths?

How would I write a computer program to calculate (find a and b) 6 = a^3 + b^3 where a and b can be any integer (positive or negative)?

I desperately need solution to this equation and I'm too noob at this stuff...

for a=1:step_size:N
    b = (6-a^3)^(1/3);
end

On April 02 2016 00:46 tofucake wrote:
this is a relatively simple process:

you have
x = a^3 + b^3

so to find b you move a over and take the cube root:

b = cube_root(x - a^3)

now you plug in x = 6 and just pick a number for a, say -3:

b = cube_root(6 - a^3) = cube_root(6 - -27) = cube_root(33) ~ 3.2

On April 02 2016 01:01 Manit0u wrote:


The thing is, a and b HAVE to be integers. I guess I won't solve it since no one has so far

The full thing to calculate is 33 = a^3 + b^3 + c^3 (people are now at trying solutions involving integers on scale of 10^14) but I thought if I simplified it to 6 = a^3 + b^3 it could potentially be easier and my computer could handle it

0^3 = 0
(+-1)^3 = +-1
(+-2)^3 = +-8
(+-3)^3 = +-27
(+-4)^3 = +-64

On April 02 2016 01:25 spinesheath wrote:

I'm pretty sure there is no solution for 6 = a^3 + b^3 for integers a and b.

Reason for the case where a is positive and b is negative or vice versa:

0^3 = 0
(+-1)^3 = +-1
(+-2)^3 = +-8
(+-3)^3 = +-27
(+-4)^3 = +-64

It's obvious that the distance between 2 rows gets larger each time. So if we don't find a solution in the first few rows, by the time the distance is greater than 6, we can't find a solution at all. Which happens to be the case rather quickly.
That's because once we have a^3 calculated, we need a -(b^3) that is exactly 6 smaller than a^3. But if the closest -(b^3) is already 37 less than a^3, the further away ones certainly won't do.

To extend on this: if you were to replace the 6 with an arbitrary x, you could just start enumerating all the possible values for a^3 until the distance between two such values becomes larger than x. Then you just do a fairly simple search for a pair that is exactly x apart.

For example if x was 26, we would get the above list. 64 is already too large (distance to 27 is 37) so we stop there. Then we start at the 27 and look among the smaller numbers for the right match, which is the 1. So 26 = 3^3 + (-1)^3. If the 27 wouldn't work out, we would work our way down. Though obviously that would be pointless in this case.

With a and b having the same sign, things should work out similarly. We can restrict that to a, b and x all being positive because of logic and reasons. So now you enumerate all a^3 until the result is larger than x and look for a match. Since everything is positive, no larger numbers can work out.

On April 02 2016 01:29 Acrofales wrote:
Excellent point. So for c=3 there is no solution! Well proved! Now manit0u can move on to c=4, 5, ...

33 = a^3 + b^3 + c^3

On April 02 2016 01:25 spinesheath wrote:
Now, if you're trying to solve 33 = a^3 + b^3 + c^3, or even worse x = a^3 + b^3 + c^3, then scratch all of the above. With just a and b we were able to determine a simple criterion for when to stop iterating. That same criterion won't work for a, b and c. Maybe there's another upper limit, but it would likely be much higher.

On April 02 2016 01:34 Blitzkrieg0 wrote:


There is no upper limit because the third value is arbitrary. If I set c to be 100 then the difference would need to be 1000033, but there is nothing stopping me from setting c to 1000 or 10000 to make the difference even larger.

On April 02 2016 01:39 spinesheath wrote:

But can you find a and b such that their cubes they add up to a value that is "in the right range" of c^3? Maybe the numbers grow apart too much and you just can't possibly hit that sweet spot. I can't reason for either way off the top of my head.

On April 02 2016 01:32 spinesheath wrote:

It's not c = 3. You misread that. His general case is this:

33 = a^3 + b^3 + c^3

So it's always cubic, but one extra term.

Also I updated my post above for the general case.

On April 02 2016 02:18 Acrofales wrote:

Exactly.

But a^3 + b^3 = 6 would find you the solution to the more general problem for c=3.

You just proved that for c=3 there is no solution. By setting c=4, you have to retry and show there is no solution such that a^3 + b^3 = -53

And so forth for all c in Z

On April 02 2016 02:46 Manit0u wrote:
I always get spooked by such stuff...

On April 02 2016 03:06 spinesheath wrote:

That parametric solution for 1... why go to such lengths when you can just use 1 = 1^3 + x^3 + (-x)^3 and still get infinitely many solutions? The same holds true for every single k^3 = k^3 + x^3 + (-x)^3.

Anyways, even though he didn't say so specifically, we can safely assume that there is no easy way to find an upper limit for the 33 case. Or else someone would already have found a solution.

Integer math is surprisingly hard, even though the numbers seem so much simpler than real numbers. In fact many problems are a lot easier if you're looking for solutions in the real numbers instead of integer solutions.

Long story short: if you want to solve this, you should probably head to university and delve into the field of discrete mathematics. It seems like people with way more experience and knowledge in the field than us combined have tried to find a solution.