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

[HelpCube]

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.

 

[Pete the Geek]

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.

 

[Sa967St]

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.

 

[vcuber13]

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)

 

[HelpCube]

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?

 

[~Adam~]

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.

 

[Timothy Ng]

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.

 

[Sa967St]

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

 

[Georgeanderre]

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.

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

 

[onlyleftname]

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

That's as far as I got...

 

[Georgeanderre]

#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

 

[HelpCube]

#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

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

 

[blah]

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 \( 8 \mid n \).