[Grade 10th Maths]

[fiery berry]

Prove that with any way of dividing natural numbers from 1-124 into two groups, there always exist 2 natural numbers in the same group whose sum is the cube of a natural number.

Help please :woeisme: I've spent 2 hours on this, but still stuck with no way out. Thank you in advance:'< (Idk what type of math this is, so I chose other)

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[coral sable]

This sounds like a pigeonhole principle problem. Basically, what you want to show is that there's some cycle of numbers where, like, a + b is a cube, b + c is a cube, and c + a is a cube.

[slim moon]

note that 125 is a cube

so we have that 1+124 = cube, 2+123 = cube, …

now if we assume a grouping without cubes is possible, then this greatly restricts our choices

[fiery berry]

Ohh yeah that makes so much more sense than what I've come up with 🥹

[fiery berry]

yup I agree, but still I can't legitimately prove it 😭

[fiery berry]

omg this brought me a whole new perspective

[coral sable]

It's actually not a pigeonhole principle problem exactly, but the rest of what I said is correct.

[slim moon]

i think a very similar line of reasoning applies

[fiery berry]

would you mind explaining further? I understand what you said but now I'm having trouble getting passed the next step 😭

[slim moon]

honestly, i would suggest trying to find three numbers such that any two of them added will sum to a cube

but i don’t know if that’s possible

[fiery berry]

I'm not too familiar with stats (if I understand what you said correctly) so I doubt it would be the ideal approach

[slim moon]

are we certain it’s not possible?

i’m going to start constructing the groups and see if anything contradicts itself

ok?

[fiery berry]

Yes thank youu

[slim moon]

so 1 and 124 are in different sets

[fiery berry]

yeah

[slim moon]

so is 1 and 63, which means 1 and 62 are in the same set

so 1 and 2 are in different sets

so 1 and 123 are in the same set?

[fiery berry]

wait i've tried this, but it gets complicated really quickly and I don't think it's efficient

[slim moon]

yeah

i was hoping something would complicate matters quickly

i think we need to use 64 somehow

if i try all the numbers to 64 and prove 63 and 62 end up together

(which they can’t)

[fiery berry]

my work on this method so far, hope it helps somehow._.

[slim moon]

wow

[fiery berry]

oh wait

62 and 63 end up together?!

[slim moon]

what

how

i mean i can see that from your paper but

what did you do to get there

[fiery berry]

Uhm bcuz 2 + 62 = 64 and 1 + 63 = 64

so they gotta be separated

[slim moon]

so 1 is with 62 and 2 is with 63

[fiery berry]

oh yeah a hole in my thinking process 😭

[slim moon]

216 is 124+92 hmm

1 and 33 must be in different groups then

which isn’t that bad; we can still have 1 and 31 in the same set

1 | 7 26 63 124
1 62 99 118 | 7 26 63 124
1 62 99 118 | 7 26 63 98 117 124
1 8 62 92 99 118 | 2 7 26 63 98 117 124
1 25 62 92 99 118 123 | 2 7 26 63 98 117 124

this is an incomplete list

it’s interesting that consecutive numbers are showing up

[fiery berry]

Oh wowww

[slim moon]

probably because of 63 and 62

[fiery berry]

Indeed

[slim moon]

why is everything either 1 greater or 1 lesser

[fiery berry]

So it has something to do with 1?

[slim moon]

i don’t know

[slim moon]

suppose two numbers are 61 apart
they couldn’t be in different groups, could they
because we have n and n-61
and 125-n has to be in one of those groups

so what if we take the two numbers 61 apart

wait

no

this fails if 125-n = n-61

but this is good because we have that n and n-61 are almost always in the same group

i’m going to hate 10th grade after this

[slim moon]

32 and 93 are the only pair that don’t have to follow this rule

[fiery berry]

You are not yet in 10th grade? I thought you're some university level math degree

[slim moon]

i’m in high school

so i’m not sure why i can’t solve this

[fiery berry]

It doesn't require knowledge higher than 9th grade, it says

[slim moon]

i know it doesn’t

this is the special type of math that doesn’t need fancy theorems to be hard

we call it olympiad

usually they at least have the decency to be nice

[fiery berry]

thank you for your help, I gotta go to sleep now, when I wake up i'll continue to think about what you've said here

[slim moon]

i’ll keep thinking about this then

[fiery berry]

I hope that after some sleep i can think better 😞

[slim moon]

n and n-91 also have to be in the same pair because of 216-n

unless n = 108

[coral sable]

I figured something out. Numbers on the interval [1, 7] can make four cubes. Numbers on the interval [8, 26] can make three. Numbers on [27, 63] can make two, numbers on [64, 91] make one, and numbers on [92, 124] make two again.

Except, yeah, 4, 32, and 108 each make one less than their respective sets.

So we actually don't need to worry about the [64, 91] interval or 32, since we need a cycle, which means each element of the cycle has to have at least two cubes it can sum to.

[coral sable]

This approach only works if you only place numbers when you know where they go.

E.g. 1 on the left, then 7, 26, 63, 124 on the right, then all the numbers that sum with those to cubes on the left, etc.

[slim moon]

i noted this as well

also, i managed to place 3 down

31 | 33
31 92 | 33 94
31 92 122 | 33 94 124
1 31 92 122 | 3 33 94 124

this means 2 and 3 are on the same side

1 5 6 25 31 62 92 99 118 122 123 | 2 3 7 26 33 63 94 98 117 124

(we can also place 5 and 6 down from summing to 8)

[coral sable]

I actually don't think it's as simple as a cycle.

[fiery berry]

I noticed that n would be on the one side and n+1 would be on the other one

I made this one at school

If I get to prove this, it's gonna contradict quite nicely, isn't it?

[slim moon]

not for all n, i assume

[slim moon]

if it’s of any interest to you, i located 3

[fiery berry]

Really?!!

[slim moon]

this was my process

[fiery berry]

I do care about 3 yea

[slim moon]

i can’t see a direct contradiction but it does feel like progress

[fiery berry]

Yes I feel like something as well

[slim moon]

@slim moon help

you can read our progress above

Okay

If we consider that 123 and 124 are in the same group then 123+124=247
So, we only need to consider the cubes which are less than 247 and more than 1

Not sure if its very relevant but just a little attemp to create a bound.

[slim moon]

they are not in the same group

[slim moon]

but this is true

we have six cubes at our disposal

[slim moon]

Can't we consider the case when they are in the same set?

well, we can, but that is impossible because they are not

Okay

But still the only cubes which we have to deal with are 8,27,64 and 125

and 216

yes

[slim moon]

5

Won't consider 1

true