Let a, b, and c be integers

But it could exist. 33, 42 and 74 where three numbers with no known solution as a sum of three integer cubes, but they found one for 74 so that leaves 42 and 33 hanging. Just a protip for anyone trying it, they found that no solution exists for any integer of magnitude smaller than (14^4)^3 so that gives you a place where to start.

>Only brainlets can't find a solution to this problem.
What does "can't find" mean? Does it mean to fathom that a solution somehow does exist? Do I have to engage in some form of double-think?

Also no solution exists in reality so what you're really saying is that stupidity is intelligence.

How could you prove that no solution exists?

>14^4
Sorry phoneposting and I messed it up, it's 10^14

Are you kidding me?

A solution exists for every integer < 1000 except for 33 and 42, evidence suggests that the solution exists but hasn't been found. Why would those 2 numbers be the only ones that cannot be written as a sum of 3 integer cubes? If you're going to make such a bold claim you have to prove it.
arxiv.org/abs/1604.07746
See? I back my claims with evidence so I suggest that you do the same, after all the backbone of math is rigor.

>A solution exists for every integer < 1000 except for 33 and 42

So you're trying to contradict me by saying that what I said was accurate? OP clearly said 33, not the general case.

What the fuck? All I'm saying is that unless you prove that a solution cannot exist for 33 then you cannot state that a solution doesn't exist.
You literally said that no solution exists in reality, which unless you can prove it it's an invalid claim. The solution most likely exists and hasn't been found.

Its very easy to prove by checking every case

a,b,c can be a 0,postive or negative integer

3 -
3 zeros
-Trivial

1+, 2 zeros
2+, 1 zero
3+
-Obviously impossible, by simple inspection

1+,1-,1zero
2+, 1-
1+, 2-

Im too lazy to do this one, but the universe of possible number is quite is small, we know that the distance of a n value and the next cannot be higher than 66 (or they give 34

It's conjectured by people smarter than you and me that a solution does exist.

It'll just be something ridiculous.

>Im too lazy to do this one, but the universe of possible number is quite is small
I don't see how you reached that conclusion.

There are no solutions for:

[math]a^3 + b^3+ c^3 = x * 9 + 4[/math]
[math]a^3 + b^3 + c^3 = x * 9 + 5[/math]

When a, b, c, and x, are integers, and that's something that's been proven, but yeah, there's nothing to suggest 33 can't be represented as the sum of three cubed integers.

>t.brainlets

The difference between the cubes of n and n+1 are always increasing as n increase

If the difference between a number of a,b,c and the rest two is higher than 33 then that number cannot be a candidate of solution, that killmost of the integers, except [-5,5], then brute force those combination yourselves faggots

>I'll just keep repeating the same thing, maybe this time I'm right
you're the third worst kind of poster

You still dont say why is wrong though

>If the difference between a number of a,b,c and the rest two is higher than 33 then that number cannot be a candidate of solution
Prove it.

>I'll say something retarded and ask them to prove me wrong
why don't you try some more numbers and convince yourself that three large cubes can add up to small numbers? takes literally 2 minutes

>You still dont say why is wrong though
That'd be showing you pity you don't deserve
You can VERY easily find out why and how exactly you are wrong by yourself

[math]\emptyset[/math]

There’s no solution for any number less than 1729 for that those a,b,c

That's very very wrong. Just google Diophantine sums of three cubes. Lots of numbers smaller than the 1000 have been found to have a solution.

lmao that was for the case when 1 number was positive and 2 number are negative

its mere algebra, taking the sign off b and c, a^3 - (b^3 + c^3) = 33, so that difference must be 33 so no solution exists if its >33

lol k, you're wrong though and have 0 examples

Eh, there’s still no integer solutions to this.

We'll not to 33, yet. But here's 30 so it proves your statement to be false by counter example. A diophantine solution to a number smaller than 1200.

None that have been found. But there's no reason to believe there aren't any.

>I'm really really braindead please spoonfeed me
-95 47 91 : 19
-95 91 47 : 19
-94 -48 98 : 16
-94 98 -48 : 16
-89 41 86 : 8
-89 86 41 : 8
-86 28 85 : 21
-86 85 28 : 21
-77 26 76 : 19
-77 76 26 : 19
-75 40 71 : 36
-75 71 40 : 36
-60 22 59 : 27
-60 59 22 : 27
-59 31 56 : 28
-59 56 31 : 28
-58 -43 65 : 6
-58 65 -43 : 6
-56 21 55 : 20
-56 37 50 : 37
-56 50 37 : 37
-56 55 21 : 20
-52 20 51 : 43
-52 25 50 : 17
-52 50 25 : 17
literally 2 minutes

>no 33

into the trash it goes :^)

>no unicorn a have been found but there’s no reason to believe there aren’t any
This is what happens when you’re all math and no science.

You brainlets haven't found a solution, I see.

You're quite retarded, don't you?

t. engineer

better than all the /pol/ and popsci threads though

Let a= 2

a^3=8 Let b= 3

b^3=27 Let c= -1.25992104989

c^3= -2

SOLUTION: -2+8+27=33

That's no diophantine solution my dude

>Let c be an integer
>c = -1.25992104989
Riiiiight,

I find it hard to believe you can't solve this problem by brute force alone. Well, it is still possible that there isn't any solution.

>I find it hard to believe you can't solve this problem by brute force alone.
I'm tempted to just it a go, but I'm sure people with vastly more resources than me have tried and failed.

Trivially untrue. If a,b,and c are restricted to the integers the maximum possible combination is a=3, b=1, and c=1. Any greater and the sum is strictly greater than 33

What's the fastest way to solve problems of this type?

Galois theory

retard

Integers can be negative senpai.

I've got a decent CPU, what would be the best way to brute force this, Veeky Forums?

If someone is willing to write an efficient script, all of us brainlets can run it and solve it together.

Why do you call him a retard for actually attempting a problem? He failed but atleast he tried like few people in this thread.

does it matter if the outcome of the solution is an integer such as -2, or does it first have to be one and only a solution?

the number c^3 is -2 which infact is an integer, how ever the number c is a decimal not an integer, however in this problem we are discussing the number c^3 remember not the number c, so would the solution now be correct or prove to be false?

this reply goes to this post

Here's some info on algorithms being used right now.
ams.org/journals/mcom/2009-78-266/S0025-5718-08-02168-6/S0025-5718-08-02168-6.pdf
All you'd have to do is implement them and keep your CPU running.

r e t a r d

You know reading that paper I can only imagine a bunch of phds getting triggered by the academical equivalent of a shitpost and trying their best to prove they are not brainlets. What a waste of effort.

>ams.org/journals/mcom/2009-78-266/S0025-5718-08-02168-6/S0025-5718-08-02168-6.pdf
>Unfortunately, we still do not know of any solution for n = 33
Authors confirmed for brainlets.

i n t e l l i g e n t

p e r s o n

youtube.com/watch?v=wymmCdLdPvM

Chai!

Ah, you guys don't seem to understand: the solution is right under your eyes. The only thing left to do is to find it. But like all things, it's in the last place you look.

Only brainlets watch a memephile episode. Big brains read a paper.

this

its probably in the island of non-trivial zeroes outside the 1/2Re line of RH

the abc conjecture being true means that there must be a solution. either our mathematics is closed with respect to itself or it isnt. the return to monoidal charactertistics implies that its all just a ticking machine. were one step closer to god boyos.

praise mochizuki

Residues of n^3 mod 7: 1,1,-1,1,-1,-1

Residue of 33 mod 7: 5

Numbers that can be written with x+y+z in {-1,1} mod 7: {4,6,1,3}

Therefore there are no such integers.

Cute, it took a minute or so to find the flaw.

Are there examples of real world benefit from discovering these numbers?

STILL no solution.

Christ, lads, you all are mega brainlets.

What? OP said
>Only brainlets can't find a solution to this problem.
I *can* find a solution, applying Elkies' method and running the algorithm for months. I just can't be bothered to do it. I'm not a brainlet because I can find a solution, I just don't want to.

How about this?

Residues of n^3 mod 9: 0,1,-1,0,1,-1,0,1,-1

Residue of 33 mod 9: 6

Therefore all cubes must be -1 mod 9, and hence of the form 3m_j + 2

I.e. (3m_1 + 2)^3 + (3m_2 + 2)^3 + (3m_3 + 2)^3 = 33

Now any cube can be written as a sum of hexagonal numbers:

n^3 = sum(j=1 to n) (3j*(j-1) + 1)

Thus we can rewrite the sum as:

(+/-) sum(j=1 to 3m_1 + 2) (3j*(j-1) + 1)
(+/-) sum(j=1 to 3m_2 + 2) (3j*(j-1) + 1)
(+/-)sum(j=1 to 3m_3 + 2) (3j*(j-1) + 1)
= 33

Which can be rewritten as (separating the +1 in the sum):

(+/-)sum(j=1 to 3m_1 + 2) (3j*(j-1))

(+/-) (3m_1 + 2)

(+/-)sum(j=1 to 3m_2 + 2) (3j*(j-1))

(+/-) (3m_2 + 2)

(+/-)sum(j=1 to 3m_3 + 2) (3j*(j-1))

(+/-) (3m_3 + 2)

= 33

Now each of the sums is divisible by three (so we denote each as 3h_t), and we also know that not all cubes can be negative or positive, thus two are negative or two are positive. Thus:

(+/-) 3h_1 (+/-) (3m_1)
(+/-) 3h_2 (+/-) (3m_2)
(+/-) 3h_3 (+/-) (3m_3)
(+/-) 2
= 33

Therefore, for some positive or negative integer k:

3k = {31,35}

Which is a contradiction since the RHS isn't divisible by three.

I tried a few on paper but this would take weeks so I give up good night.

You just have to make an ice-cube machine which makes cubes of different sizes. Then put them in cups and come back when they have melted and see if the total volume is 33 ;)

Alright, I'll bite.
How do you make an ice cube of negative volume, then?

You make the ice-cubes out of coke and the negative ones have menthos inside them

Or use anti-matter