Why is 2^n + 3^n never a perfect cube? | JBMO Shortlist

[@andreben6224]

I concluded a bit differently 😄 So when we found that n=3k (via mod 6 congruence), I thought of factoring the expression: if 2^n + 3^n = x^3 then 2^(3k)=( x-3^m )( x²+x*3^m+3^(2m) ) since n is bigger than 2, we know that x is odd and x²+x*3^m+3^(2m) is also odd but strictly larger than 3, thus the RHS can't be a power of 2. I really like that bounding the expression strictly between two consecutive cubes. So beautiful !!

[@laplace1139]

oh so you have a prime factor contradiction, nice

[@cobokobo2115]

amazing .....kzbin.info/www/bejne/rZO0nntnn6l-i6M

[@mr.kaiden7159]

How about n=3k±1 ?

[@richardfredlund8846]

once n is a multiple of 3 we could use fermats last theorem, to say that the result wont be a cube but the direct proof based on consecutive cubes is much more satisfying.

[@johnnath4137]

Very nice! (I worked out - by hsnd! - all the cubic residues mod 19, and found that the residues of 2^n + 3^n (mod 19) were not among them, but I like your method much better).

[@leif1075]

Isn't there a way to prove without modular bullshit??

[@johnnath4137]

@@leif1075 I suppose it might be possible to assume that it is a perfect cube for some n, and then derive a contradiction, but I haven’t been able to make any progress along these lines.

[@leif1075]

@@johnnath4137 How would most ppl solve this then? Because I don't think most ppl would think of modular arithmetic..

[@riadsouissi]

Did same with mod 19. I just had to try before giving up. 19 started to be quite a lot of computation :)

[@johnnath4137]

@@leif1075 Aonother possible method is tto try to show that the expression lies between two consecitive cubes. Induction is another tool. For these number theory problems, one has to have recourse to all the tools at one's disposal.

[@copiryte9535]

assume that fermat's last theorem is trivial and is left an exercise for the reader.

[@cobokobo2115]

amazing .......kzbin.info/www/bejne/rZO0nntnn6l-i6M

[@christopherstokes9393]

Fortunately, you only need the special case n = 3, which was proven by Euler... the proof is still not trivial, but it's doable :P .

[@Lorenz495]

Instead of the last step, couldn't you just say, that (2^n)^3+(3^n)^3 = m^3 for a positive integer m, because of Fermat's last theorem?

[@eduardomalacarne9024]

That is true by the condition n might be bigger than 0

[@cobokobo2115]

amazing .....kzbin.info/www/bejne/rZO0nntnn6l-i6M

[@oliveirafilipe5416]

Then it works, since n is natural

[@Horinius]

You should have used a different letter other than n when you wrote n = 3N + r to *avoid confusion*

[@mrityunjaykumar4202]

why (2,9) does not follow fermat's little theorem.. that is.. gcd(2,9)=1 => 2^(9-1)=1 (mod 9).. but 2^8 is actually 4(mod 9)

[@FaranAiki]

Of course, 9 isn't a prime. 9 = 3 x 3. You may check euler-totient function though. phi(9) = 6, and therefore 2^6 is congruent to 1 (mod 9)

[@mcwulf25]

I played with this a while but didn't pick up on the inequality solution.

[@cobokobo2115]

amazing .......kzbin.info/www/bejne/rZO0nntnn6l-i6M

[@6388-s2n]

can you solve this with a graph. for x

[@tianqilong8366]

i did not see a way using graph to solve there. you can try to draw it in desmos, there is no pattern there

[@cobokobo2115]

amazing .......kzbin.info/www/bejne/rZO0nntnn6l-i6M

[@jimmykitty]

Oh I need to learn the Chinese Remainder Theorem 😺

[@johnnath4137]

You are not allowed to - you would risk getting sanctioned!

[@jimmykitty]

@@johnnath4137 why?

[@johnnath4137]

@@jimmykitty Haven't you been listening to the news? The White House has sanctioned most things emanating from China 😺

[@jimmykitty]

@@johnnath4137 Oh! I didn't know about this before 😸

[@cobokobo2115]

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[@gmchess7367]

Really like the solution

[@cobokobo2115]

amazing .......kzbin.info/www/bejne/rZO0nntnn6l-i6M

[@patrick-8068]

Cube is​ ax+b​ lesser 0. log(2) n​ + log(3) n​ lesser​ 0 Form... 2an+1.3bn lesser​ 0 law​ cube​ is​ ax​ is​ R.​ both​ bx is​ R2.​ Answer.

[@OndrejPopp]

2^n + 3^n = (2 + 3)^n = 5^n = 5^3 for n = 3. QED. Naah... just kidding! 🙃

[@mariomestre7490]

Genial, merci

[@adokoka]

A minor detail, but those things are important in mathematics. At 8:25, don’t forget the case when n = 0, as n is a member of N not N*

[@晓阳-d3p]

1）mod 2）check is it be between two consecutive cube

[@h4ck314]

How do you prove 2) ?

[@cobokobo2115]

amazing .......kzbin.info/www/bejne/rZO0nntnn6l-i6M

[@himu1901]

In the beginning why did u take only p = 9?

[@4hmqd]

because n^3 is 0, 1 or -1 mod 9 and 3^n is always 0 for n>2, but I think everything here also works if you take 4

[@himu1901]

@@4hmqd yeah thanks

[@cobokobo2115]

amazing .......kzbin.info/www/bejne/rZO0nntnn6l-i6M

[@mohanabavi]

Why n = 3k? In fact r was either 1 or 2.

[@petersievert6830]

It's a bit confusing. His residues mod 9 start at n=2 . -1 and 1 are reached at n=3,6,9,...

[@cobokobo2115]

amazing .......kzbin.info/www/bejne/rZO0nntnn6l-i6M

[@shigototravaillez9972]

Question - can’t you just: Show n=0 -> 2 which is not a cube Assume it works: 2^n + 3^n = 3^(n+m), m is integer Divide both sides by 3^n (2/3)^n +1 = 3^m (2/3)^n is not an integer unless n=0 (and n=0 doesn’t solve), and is greater than 1 3^m is either an integer, or a fraction less than 1 In both cases, the equality doesn’t hold, so no cubes.

[@n.rv..n]

A cube is of the form x^3, not 3^x

[@leif1075]

Fuck Modular Madness..surely you can solve this with just algebra or another way

[@obrod7080]

What's wrong with modular arithmetic

[@cobokobo2115]

amazing .......kzbin.info/www/bejne/rZO0nntnn6l-i6M

[@cobokobo2115]

@@obrod7080 amazing .......kzbin.info/www/bejne/rZO0nntnn6l-i6M

[@FaranAiki]

"Where there is an integer, there is a modular madness."