# A^3 + b^3 + c^2 = 2012

1. Jun 8, 2012

### Lilt

1. The problem statement, all variables and given/known data

a^3 + b^3 + c^2 = 2012 ; a, b, and c are positive integers.

2. Relevant equations

I was thinking prime factorization, but that was a bust.

3. The attempt at a solution

2012 has a beast of a prime in it (503) so I have no idea how to solve this theoretically. Help.

In retrospect, factorization isn't a key element here anyway.

2. Jun 8, 2012

### Dickfore

I think there is no solution.

3. Jun 8, 2012

### Villyer

I made a quick brute-force program that didn't return any solutions either.

4. Jun 9, 2012

### D H

Staff Emeritus
Yes, there is a solution.

5. Jun 9, 2012

### skiller

I think you must have written it a bit too quickly then. :tongue:

6. Jun 9, 2012

### algebrat

a and b are less then 12, since $\sqrt[3]{2012}<13$. So try things like 123+13 et cetera.

7. Jun 9, 2012

### algebrat

is this from a class and/or textbook? are there any relevant methods we should know about? give us some more context, this may use an obscure method (to avoid checking so many cases).

8. Jun 9, 2012

### D H

Staff Emeritus
With modern computers, it's not all that many cases. A simple brute force program in python or perl takes but a handful of lines of code and will find the solutions ASAP. A couple of hands full are needed with a more traditional language such as C or C++. Excel could also be used as a brute force tool.

Aside from brute force or some refinements to it (but why bother?), some obscure number theoretical methods are most likely needed.

Then again, number theory has never quite grabbed me. Perhaps those methods aren't so obscure to someone who knows the subject.

9. Jun 9, 2012

### Dickfore

Ah, I just noticed I made a mess with the constraints, and I missed the solutions. After revising it, I found the two possible solutions.

I say two, because if (a, b, c) is a solution, then so is (b, a, c). But, in searching for a solution, it is sufficient to assume that $a \ge b$. Then, you find one solution.

Let us look at the constraints for a and b. Because c is a positive integer, we have $c \ge 1 \Rightarrow c^2 \ge 1 \Rightarrow a^3 + b^3 = 2012 - c^2 \le 2011$. Then, suppose $a$ is fixed. We have the following constraints for b:
$$b \ge 1, \ b \le a, \ b^3 \le 2011 - a^3 \Leftrightarrow b \le \sqrt[3]{2011 - a^3}$$
Notice that there are two constraints for the upper bound of b. The smaller of them determines the true upper bound:
$$1 \le b \le \min \left\lbrace a, \sqrt[3]{2011 - a^3} \right\rbrace$$

Of course, the minimum is no greater than any of its arguments. This establishes the bounds on a:
$$\left \lbrace \begin{array}{rcl} 1 & \le & a \\ 1 & \le & 2011 - a^3 \end{array}\right. \Leftrightarrow 1 \le a \le \sqrt[3]{2010}$$

Notice that $\sqrt[3]{2010} \approx 12.62$, so $1 \le a \le 12$ are the integer bounds for a.

Now, you may make a double loop to go through all allowed values for a and b. For a given pair (a, b), $c^2 = 2012 - a^3 - b^3$ is uniquely determined. But, is this a true solution when c is an integer? This establishes a selection criterion.

The program spits out a unique positive integer triple (a, b, c) as a solution. Due to the above mentioned symmetry, (b, a, c) is also a solution.

You ought to have enough information to sift through all the cases. If you are patient, you may even do it by hand. We are not allowed to give you a detailed solution, nor post the final answer without you making a significant attempt, so this is where I stop.

Thanks D_H for re-checking.

10. Jun 9, 2012

### Villyer

Ah, I read $a^{2} + b^{2} + c^{2} = 2012$

I apoligize for giving a false statement.

I edited the program, adjusted restraints on a and b to make it run faster, and it returned two possibilities: one being in the form a,b,c and the other being b,a,c.

I wouldn't know how to go about solving this without a program, however, and I would be interested to see if there was a simple way to do this.

Last edited: Jun 9, 2012
11. Jun 9, 2012

### Lilt

Wow, thank you so much, but your methods went a bit over my head. Could you suggest which class or material I would need to study to be capable of working through this as efficiently as you just did? I am not very strong with calculator programs, and I was hoping this problem wouldn't mandate that, but it seems the time has come for me to learn them well.

As would I, but now I think I need to find something like a calculator workshop lol.

Last edited: Jun 9, 2012
12. Jun 9, 2012

### algebrat

For precalculus, the method would be obscure.

We still have not gotten confirmation of class, text, context or expectation in technique.

Last edited: Jun 9, 2012
13. Jun 9, 2012

### ehild

Hi Lilt,

If you are not expected to be familiar with programming, you can solve this problem by trial and error. If you have luck, it will not last too long. You know that neither a nor b can be greater than 12.

Let be a≥b. Fist assume that a=12. Its cube is 1728, so b3+c2≤284, so b can not be greater than 6. If you try to subtract the cube of b= 6, 5, ...1 neither of them results in a square of an integer.

So try a=11 and repeat the procedure. It is the same as Dickfore's program but you are the computer. :tongue2:

ehild

14. Jun 12, 2012

### ViktorS

For A = 2 --> 7≤B
For A = 3 --> 7≤B
For A = 4 --> 7≤B
For A = 5 --> 6≤B
For A = 6 --> 5≤B

... You can already cancel out some options like this.

15. Jun 13, 2012

### Ratch

16. Jun 13, 2012