# a^3 + b^3 = 22c^3



## HelpCube (Jan 21, 2012)

```
a^3 + b^3 = 22c^3
```

My friends and I have been trying to solve this for a few weeks now, and have gotten almost nowhere. a, b, and c are the variables and they are all positive whole numbers. Any help? (no answers for now )

I've figured out a good method of guess and check, but the answers could be in the thousands for all I know. My guess and check method is adding perfect cubes whose sum of their cube roots are multiples of 22. For example, if you add 56 cubed and 10 cubed that number is divisible by 22, and if you get a perfect cube after dividing it by 22 you found the answer. I've gotten through 88, and I don't think I'm going to hit the answer any time soon.


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## Pete the Geek (Jan 21, 2012)

HelpCube said:


> ```
> a^3 + b^3 = 22c^3
> ```
> 
> My friends and I have been trying to solve this for a few weeks now, and have gotten almost nowhere. a, b, and c are the variables and they are all positive whole numbers. Any help?


I'm not giving you the answer, but I suggest you read this article about Fermat's Last Theorem.


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## Sa967St (Jan 21, 2012)

The values of a, b and c are all well in the thousands. The first person to ever solve this problem used guess and check and it took him over a year. 
Interestingly, a and b are prime numbers.


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## vcuber13 (Jan 21, 2012)

Pete the Geek said:


> I'm not giving you the answer, but I suggest you read this article about Fermat's Last Theorem.


 
this has a coefficient though

if you only want the answer:


Spoiler



17,299^3 + 25,469^3 = 22(9,954^3)


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## HelpCube (Jan 21, 2012)

Sa967St said:


> The values of a, b and c are all well in the thousands. The first person to ever solve this problem used guess and check and it took him over a year.
> Interestingly, a and b are prime numbers.


 
Wow. Is there a better method of guess and check than the one I am using?


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## ~Adam~ (Jan 21, 2012)

Sa967St said:


> The first person to ever solve this problem used guess and check and it took him over a year.


 
Any idea what year this was?

Seems like a waste of a year to me when you can now write a simple program to guess and check for you.


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## Timothy Ng (Jan 21, 2012)

*Try these algebra problems!!!!!*

Hi guys,

I read this thread because it reminded me of the Enrichment problems i did when i was in year 9. This problem seems really hard, i find it interesting.

Then i looked back of the problems i did, involving algebra, and these are a few i can remember. Please try them!!!

1. Prove that a number 10^(3n+1), where n is a positive integer, cannot be represented as the sum of two cubes of positive integers. 

2. Let X and Y be positive integers such that X^2+3X+y^2=404. What is the value of X+Y^3? Find all possible solutions.

3. A positive integer n is such that numbers 2n+1 and 3n+1 are perfect squares. Prove that n is divisible by 8.


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## Sa967St (Jan 21, 2012)

cube-o-holic said:


> Any idea what year this was?
> 
> Seems like a waste of a year to me when you can now write a simple program to guess and check for you.


 
In the 1940s.
http://www.rhinocerus.net/forum/lang-rexx/175236-3-b-3-22-c-3-a.html


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## Georgeanderre (Jan 21, 2012)

Timothy Ng said:


> Hi guys,
> 
> I read this thread because it reminded me of the Enrichment problems i did when i was in year 9. This problem seems really hard, i find it interesting.
> 
> ...


 
The first one is a standard AS level question, comes up biannually in C2

the second and third would be substitution or induction... a li'l harder


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## onlyleftname (Jan 21, 2012)

c = -(-1/22)^(1/3) (a^3+b^3)^(1/3)

That's as far as I got...


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## Georgeanderre (Jan 21, 2012)

#2 -

```
X^2+3X+y^2 = 404

Y^2 = -X^2-3X+404
Y^2 = (-X^2-3X+404)(-X^2-3X+404)
Y^2 = X^2+3X^3-404X^2- ...

x^4+6X^3-799X^2-2424X+163216

y = sqrt(x^4+6 X^3-799 X^2-2424 X+163216)
hence,

sqrt(x^4+6 X^3-799 X^2-2424 X+163216)sqrt(x^4+6 X^3-799 X^2-2424 X+163216) = x^4+6 X^3-799 X^2-2424 X+163216
```

ye.. :fp that failed, better luck next time


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## HelpCube (Jan 21, 2012)

Georgeanderre said:


> #2 -
> 
> ```
> X^2+3X+y^2 = 404
> ...


 
Lol nice. I'll try one of those in a bit.


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## blah (Jan 21, 2012)

2.

```
$ perl -e 'for $x (1..20){for $y (1..20){print(($x+$y**3)." ") if $x**2+3*$x+$y**2==404}}'
8001 2757 1016 529
```

3.
Since \( (4k+1)^2 \equiv (4k+3)^2 \equiv 1 \pmod 8 \), we have that any odd perfect square \( \equiv 1 \pmod 8 \), from which it follows that \( 2n+1 \equiv 1 \pmod 8 \), implying that \( 4 \mid n \), which means \( 3n+1 \) is odd, hence \( 3n+1 \equiv 1 \pmod 8 \), so \( 8 \mid n \).


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