Current Events > A clever math question for you 140+ IQ guys out there.

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FLOUR
06/02/23 4:19:19 PM
#1:


Prove whether or not there are positive integers x, y, z such that x^3 - y^3 = 2^z .

All that's needed to solve this is advanced algebra and a bit of ingenuity.

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Flauros
06/02/23 4:20:06 PM
#2:


do your own homework

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coolguyjimmy
06/02/23 4:21:48 PM
#3:


Flauros posted...
do your own homework

Intelligence would be solving the question. Wisdom is realizing that it's a homework question and not answering.
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Bass
06/02/23 5:00:39 PM
#4:


Just ask ChatGPT.

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Questionmarktarius
06/02/23 5:04:11 PM
#5:


coolguyjimmy posted...
Intelligence would be solving the question. Wisdom is realizing that it's a homework question and not answering.
Charisma is getting someone else to do your homework.
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Agonized_rufous
06/02/23 5:07:30 PM
#6:


Set them all to 0 and say 0 is positive

Done

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Turbam
06/02/23 5:08:03 PM
#7:


Agonized_rufous posted...
Set them all to 0 and say 0 is positive

Done
Based

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Questionmarktarius
06/02/23 5:08:31 PM
#8:


Agonized_rufous posted...
Set them all to 0 and say 0 is positive
2^0 is 1.
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Medussa
06/02/23 5:09:07 PM
#9:


Agonized_rufous posted...
Set them all to 0 and say 0 is positive

Done

0 != 1, though.

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Agonized_rufous
06/02/23 5:10:58 PM
#10:


Yes but it'll show effort and he'll get half credit

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g980
06/02/23 5:37:18 PM
#11:


didnt you make this topic yesterday about your number theory homework

this feels like either it wants a proof by induction, or it's one of those million dollar unsolvable math problems and youre trolling

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eggcorn
06/02/23 5:41:37 PM
#12:


coolguyjimmy posted...
Intelligence would be solving the question. Wisdom is realizing that it's a homework question and not answering.
Got em

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CoyoteTheGreat
06/02/23 5:42:24 PM
#13:


Flauros posted...
do your own homework

I haven't heard this one in at least 15 years, XD. What a blast from the past.

Nowadays it should be "Get an AI to do your homework for you" though.

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Zonbei
06/02/23 5:53:35 PM
#14:


FLOUR posted...
Prove whether or not there are positive integers x, y, z such that x^3 - y^3 = 2^z .

All that's needed to solve this is advanced algebra and a bit of ingenuity.

IQ doesnt really correlate with math skills. Someone doesnt need to be able to do this homework problem for you in order to have a high IQ.

Also, IQ is meaningless as a measure of intelligence.

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PrettyBoyFloyd
06/02/23 6:22:46 PM
#15:


What does it do?

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UnholyMudcrab
06/02/23 6:24:45 PM
#16:


Calculus is hard

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NeonTentacles
06/02/23 6:25:20 PM
#17:


I'm 138 and only took up to Calculus 5, so I cant. Sorry >_>

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Questionmarktarius
06/03/23 12:37:47 AM
#18:


You just need two cubes where the difference is a power of two.
The easiest solution would be a cube that's one greater than any given power of two. Given that numbers are infinite, there's probably a collision somewhere.

X^3 = 2^Z + 1
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Eviora
06/03/23 9:16:30 AM
#19:


There is no solution, but unless TC has a considerably more elegant proof than I do, this is an infinite descent problem, so the proof isn't as trivial as they make it sound.

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BCAM1984
06/03/23 9:24:06 AM
#20:


coolguyjimmy posted...
Intelligence would be solving the question. Wisdom is realizing that it's a homework question and not answering.

LOL, this.

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Jiek_Fafn
06/03/23 9:24:48 AM
#21:


The answer is 6

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lderivedx
06/03/23 2:55:03 PM
#22:


No solutions because 2^r - 1 is not even for r>0.

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FLOUR
06/04/23 6:24:57 PM
#23:


this is an infinite descent problem, so the proof isn't as trivial as they make it sound.

That's where the IQ of 140+ comes into play. And that's the method that came to my mind for even values of x and y.


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FLOUR
06/06/23 8:04:31 PM
#24:


Pro-tips:

1) Prove that there are no odd values of x and y that satisfy the equation. (Easy part)

2) Prove that there are no values of z < 4 that satisfy the equation. (A bit tougher)

3) Assume there is a value of z > 3 that satisfies the equation and reach a contradiction. (Harder part)

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Takuya_Lee
06/06/23 8:06:36 PM
#25:


Oh so you can do your own homework TC. Good for you.

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lderivedx
06/06/23 11:46:02 PM
#26:


I don't think this requires infinite descent. You can do it with some basic algebra.

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TheMikh
06/07/23 12:02:47 AM
#27:


bruteforced all combinations of f(x,y) = 2^z from 1 to 10,000 to no avail

if i wasn't sick this would be an interesting problem to look into further

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