I think there is none. From 1-10000 they all had more than 20 (probably infinite) digits. Among the closest were:
80838083 -> square-root: 8991.00011122233275662288
99999999 -> square-root: 9999.99994999999987499999
I'm trying to find a php-command to determine if a number is element of natural numbers. Then I can tell you right away instead of scrolling through endless pages of numbers 🙂
However, no matter until what number I check, there will always be a larger number that might work. So you probably need a mathmatical proof why there isn't any number. But I don't know how to do that.
Originally posted by crazyblueI also went for a computational approach rather than a proof one because I figured that it might give a quick answer.
I think there is none. From 1-10000 they all had more than 20 (probably infinite) digits. Among the closest were:
80838083 -> square-root: 8991.00011122233275662288
99999999 -> square-root: 9999.99994999999987499999
I'm trying to find a php-command to determine if a number is element of natural numbers. Then I can tell you right away instead of scroll ...[text shortened]... robably need a mathmatical proof why there isn't any number. But I don't know how to do that.
I was working in Matlab but I think I've got one up on you in terms of determining whether a number is a repeat.
Say N is the number we are working with.
floor(N/ceil(log(N))) is the (hopefully) repeating section. And then:
if N - floor(N/ceil(log(N))) = ceil(log(N))*floor(N/ceil(log(N))) we have a repeating number.
My code will only return something if it finds such a square which satisifies this condition. Checking up to 10 million returns nothing.
Originally posted by XanthosNZi've almost got it, if all numbers of the form (10^n)+1 are square free it can't be done.
I also went for a computational approach rather than a proof one because I figured that it might give a quick answer.
I was working in Matlab but I think I've got one up on you in terms of determining whether a number is a repeat.
Say N is the number we are working with.
floor(N/ceil(log(N))) is the (hopefully) repeating section. And then:
if N - f ...[text shortened]... finds such a square which satisifies this condition. Checking up to 10 million returns nothing.
if p^2|(10^n)+1 then LET r = (10^n +1)/p^2 since 2,3,5 don't divide 10^n+1 p^2 >=49 therefore number of digits in r is less than or equal to n keep multiplying r by 4 until you get an n-digit number call it X.
repeat X is X*(10^n+1)= [(4^k)*r]*[r *(p^2)]=(2^k)^2*(r^2)*(p^2)
if 10^n+1 is always square free then repeat of an n-digit number x
is x*(10^n+1) if thats a square (10^n+1)|x but thats a contradiction because 10^n+1 {an (n+1)-digit num} > x {an n-digit num}
EDIT: i'm using a|b to mean "a divides b"
ok got it, 10 is a primitive root of {Z/(289)Z} to check that note that phi(289)=phi(17^2)=17^2-17=17*16
therfore if 10^17 != 1mod 289 and 10^16 != 1mod289 then 10 is a primitive root. therefore exists k s.t. 10^k = (-1)mod289 =>
10^k+1 = 0mod289 ergo 289|10^k+1 and 10^k+1 is not square free.
namely k=phi(289)/2 = 17*8 = 136
so let n = 256*(10^136 + 1)/289 (has 136 digits)
let r = 16* (10^136 + 1)/17 (also has 136 digits)
r^2 = (repeat n) (has 272 digits)
n = 885813148788927335640138408304498269896193771626297577854
671280276816608996539792387543252595155709342560553633217
9930795847750865051904
r = 941176470588235294117647058823529411764705882352941176470
588235294117647058823529411764705882352941176470588235294
1176470588235294117648
this was the smallest i could find
Originally posted by aginisAre you really saying n is the square root of r here? I don't understand how you found that number, but I'm impressed 🙂
ok got it, [...]
n = 885813148788927335640138408304498269896193771626297577854
671280276816608996539792387543252595155709342560553633217
9930795847750865051904
r = 941176470588235294117647058823529411764705882352941176470
588235294117647058823529411764705882352941176470588235294
1176470588235294117648
this was the smallest i could find
Originally posted by TommyCok u want to repeat a number X with n digits
I don't follow you there . . . ?
step 1 add n zeroes to the end of the number (same as multiplying X by 10^n)
step 2 add X to your result (X*10^n + X) = X * (10^n + 1)
i want that to be a square so i try X = (10^n + 1) but that has one too many digits so i'm going to divide out a square (a square divided by a square is still a square) from X = (10^n + 1)
in order to make it have less digits.
so i need to find a square that divides (10^n + 1)
so far i haven't fixed n so i can fiddle around with it using mods i figured out that 289|10^136+1 now X is too short so i multiply by 4 until i get a number of the right length (multiplying a square by a square is still a square) therfore my new X1*(10^136 + 1) is a square.
but thats equal to repeat X1 (because X1 is of length 136)
to find the root i just sqrt all my actions divide by 289 became divide by 17, and multiply by 256 became multiply by 16.
Originally posted by crazyblueno r is the square root of n repeated as in the number nn.
Are you really saying n is the square root of r here? I don't understand how you found that number, but I'm impressed 🙂
i'm kinda impressed myself i didn't think i had it in me 😵
Originally posted by aginisExcellent reasoning.
ok u want to repeat a number X with n digits
step 1 add n zeroes to the end of the number (same as multiplying X by 10^n)
step 2 add X to your result (X*10^n + X) = X * (10^n + 1)
i want that to be a square so i try X = (10^n + 1) but that has one too many digits so i'm going to divide out a square (a square divided by a square is still a square) from X = ...[text shortened]... t all my actions divide by 289 became divide by 17, and multiply by 256 became multiply by 16.
Originally posted by aginisI understand the notation, but still don't follow your logic when it comes to repeat(x)=x(1 + 10^n). The only thing I could extract was that the unit part of x could not be 2, 3, 7 or 8, if there existed a natural number m st m^2 = repeat(x).
ok u want to repeat a number X with n digits
step 1 add n zeroes to the end of the number (same as multiplying X by 10^n)
step 2 add X to your result (X*10^n + X) = X * (10^n + 1)
i want that to be a square so i try X = (10^n + 1) but that has one too many digits so i'm going to divide out a square (a square divided by a square is still a square) from X = t all my actions divide by 289 became divide by 17, and multiply by 256 became multiply by 16.
Ho hum, I did only scrape a 2.1, many years ago now . . .
Btw, congratulations on finding an answer.
Originally posted by TommyCmore later 5 AM must sleeeeeeep
I understand the notation, but still don't follow your logic when it comes to repeat(x)=x(1 + 10^n). The only thing I could extract was that the unit part of x could not be 2, 3, 7 or 8, if there existed a natural number m st m^2 = repeat(x).
Ho hum, I did only scrape a 2.1, many years ago now . . .
Btw, congratulations on finding an answer.
My solution is something like this:
Def: R is a positive number not started with a zero. Then RR is R repeated twice, we call it a 'repeat'. The number of figures in RR is always twice the number of R.
Ex: If R = 2006 then RR = 20062006.
Def: 1(n)1, when n is a non-negative number, is a number starting with a one followed by n zeroes, ending with an one.
The numbers of figures of 1(n)1 is always two more than n.
Ex: When n=3 then 1(n)1 = 1(3)1 = 10001. 1(0) = 11.
Okey here we go.
I'd like to prove that RR is never a square.
We first take R=2006 as an example. Then RR=20062006.
To make R to RR we have to multiply R with 1(4)1. 2006*10001=20062006.
That means 1(4) is a factor in 20062006.
If we divide RR with 1(4)1 then we come back to R.
If 1(4)1 is not by itself a square then 20062006 is not a square either.
If we take any R with n figures, then RR is divisible with 1(n)1.
And if 1(n)1 is not a square then RR can't be a square either.
Can 1(n) ever be a square by itself?
I say no. Here's the proof of that.
1(0)1 * 1(0)1 = 11 * 11 = 121 = 1(0)2(0)1
1(1)1 * 1(1)1 = 101 * 101 = 10201 = 1(1)2(1)1
1(2)1 * 1(2)1 = 1001 * 1001 = 1002001 = 1(2)2(2)1
and in general
1(n)1 * 1(n)1 = 1(n)2(n)1.
Since every 1(n)1 squared always has a two in the middle it can never be a square.
And that proves that a repeat can never be a square.
Originally posted by FabianFnasok first
My solution is something like this:
Def: R is a positive number not started with a zero. Then RR is R repeated twice, we call it a 'repeat'. The number of figures in RR is always twice the number of R.
Ex: If R = 2006 then RR = 20062006.
Def: 1(n)1, when n is a non-negative number, is a number starting with a one followed by n zeroes, ending with a ...[text shortened]... he middle it can never be a square.
And that proves that a repeat can never be a square.
"If we divide RR with 1(4)1 then we come back to R.
If 1(4)1 is not by itself a square then 20062006 is not a square either."
this is logic is false if you multiply two non-squares {R and 1(n)1 } you can still get a square in return. Examples. 28*63 or 3*27 or in general (p*p*k)*(q*q*k).
second your proof that 1(n)1 is never a square is incomplete, you only prove that the square root of a number of the form 1(n)1 is not of that form, what about other forms?
I prefer this method:
Let x^2 = 1(n)1
3 does not divide 1(n)1 therefore 3 does not divide x
(x-1)*(x+1)=x^2 - 1 = 1(n)0 = 10^k
if 3 does not divide x, 3 divides either (x -1) or (x+1)
but 3 does not divide 10^k = (2*5)^k therefore we have a contradiction.
Essentially that is the same as trying to find solutions for the congruence x^2~2mod3 (legendre's symbol denotes a square remainder a mod prime p as (a/p)=1 and a non-square remainder a mod p as (a/p)=-1 (and 0 if a ~ 0mod p))
since (1mod3)^2 = (1^2)mod3 = 1mod3
(2mod3)^2 = 4mod3=1mod3 there are no solutions to x^2~2mod3
since we are looking for x^2~1(n)1 mod 3~2mod3 there are no integer roots mod 3 but if "a" is a perfect square then it must have an integer root mod (p) for all p. therefore 1(n)1 is not a perfect square.
examples
1~1mod3
4~1mod3
9~0mod3
16~1mod3
25~1mod3
...
437^2 = 190969 = 3*63656+1~1mod3
finally see an example above for a number which when repeated is a square.