if n=-1 is a valid answer... n=-8/11 also works [11111=(101/121)^2] but neither 0,-1 is valid number bases,tho guess -1 is like binary. Wonder i fu could count with -8/11. n=1/3 also works (11/9)^2 depending on which coeeficients u want to get rid of.
negative bases are a thing, and rational bases are also workable. The main issue with base -1 is that there's exactly the digit 0 (so technically you can't even write 11111) and you can't really encode any numbers that way. At least with base 1 the number of 0s would give the value (if you specify the length ofc), but for -1 it just alternates between 0 and 1.
Good evening, from Brazil! In decimal base n^4+n^3+n^2+n+1 As it is a perfect square WLOG x>0 and x integer x^2=n^4+n^3+n^2+n+1; As x> n^2; x= n^2+ k; k>0 and k integer so n^4+n^3+n^2+n+1=(n^2+k)^2 f(n)=n^3-(2k+1)n^2+n-k^2+1=0 It is easy to show that f(n)=2kn^2 and n^2+n +1>k^2 Then for k>=3 there is no positive integer solution, as any positive solution n*, 2k-10, so we do not have positive integer solution. For k=2 n^3 -3n^2+n-3=0; As n*>0 and n* | -3: n* have two options n*=1 or n*=3. But for n*=1 does not work. And checking for n*=3 is good. So the only solution is n=3.
Actually -1 indeed somewhat works. 11111 would be 1 - 1 + 1 - 1 + 1 = 1 , thus a perfect square :-)
cant agree more, but I think in a lot of vids, there is an assumption that n is a positive integer
Here we are talking about base n not n as a any real number. Bases are always natural number
@@rajsingh8372 Yes, of course. Btw already mentioned in above replies.
Right. Except that a base -1 number can only have 0's as digits, according to the definition of negative base. So 11111 can't be a base -1 number.
the way to go man!
kharap
if n=-1 is a valid answer... n=-8/11 also works [11111=(101/121)^2] but neither 0,-1 is valid number bases,tho guess -1 is like binary. Wonder i fu could count with -8/11. n=1/3 also works (11/9)^2 depending on which coeeficients u want to get rid of.
negative bases are a thing, and rational bases are also workable. The main issue with base -1 is that there's exactly the digit 0 (so technically you can't even write 11111) and you can't really encode any numbers that way. At least with base 1 the number of 0s would give the value (if you specify the length ofc), but for -1 it just alternates between 0 and 1.
Good evening, from Brazil!
In decimal base n^4+n^3+n^2+n+1
As it is a perfect square WLOG x>0 and x integer
x^2=n^4+n^3+n^2+n+1; As x> n^2; x= n^2+ k; k>0 and k integer
so n^4+n^3+n^2+n+1=(n^2+k)^2
f(n)=n^3-(2k+1)n^2+n-k^2+1=0
It is easy to show that f(n)=2kn^2 and n^2+n +1>k^2
Then for k>=3 there is no positive integer solution, as any positive solution n*, 2k-10, so we do not have positive integer solution.
For k=2 n^3 -3n^2+n-3=0; As n*>0 and n* | -3: n* have two options n*=1 or n*=3.
But for n*=1 does not work. And checking for n*=3 is good.
So the only solution is n=3.
A‽₩4₩0
awesome method bruh
Sorry I broke the likes as perfect square(144)
One question tho: why 11111n means n4 + n3 + n2+ n + 1? anyone knows hahaha?
never mind, got it lol, it is called base positional number system, for those who also dont know :)
@@tianqilong8366 thanks m8
It is n base.
So it is n^4+ n^3 + n^2 + n + 1 in decimal base.
If n=10 11111=10^4+10^3^10^2+10^1+1=11111
I hope I had helped.
@@pedrojose392
amazing solution
I initially read this square to d’alembert operator😂
Get your head out of relativity, and back into number theory
@@insouciantFox 😝
And what about
When is it a prime ?
Base 10?
@@mrwiter4255 yes
answer= 1 my study try
1^4+1^3+1^2+1^1+1^0=5, which is most definitely not a perfect square
Proof?
@@fix5072 also, there is no such number as 1 in 1 number system, so 11111(1) makes no sense
Aakkasam1111111111+2222222222+786