- 05 Jul '06 23:49I 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. - 06 Jul '06 00:02 / 1 edit

I also went for a computational approach rather than a proof one because I figured that it might give a quick answer.*Originally posted by crazyblue***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. - 06 Jul '06 00:22 / 1 edit

i've almost got it, if all numbers of the form (10^n)+1 are square free it can't be done.*Originally posted by XanthosNZ***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" - 06 Jul '06 01:05 / 1 editok 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 - 06 Jul '06 01:47

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*Originally posted by aginis***ok got it, [...]**

n = 885813148788927335640138408304498269896193771626297577854

671280276816608996539792387543252595155709342560553633217

9930795847750865051904

r = 941176470588235294117647058823529411764705882352941176470

588235294117647058823529411764705882352941176470588235294

1176470588235294117648

this was the smallest i could find - 06 Jul '06 01:50

ok u want to repeat a number X with n digits*Originally posted by TommyC***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. - 06 Jul '06 01:57

Excellent reasoning.*Originally posted by aginis***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. - 06 Jul '06 02:03 / 1 edit

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).*Originally posted by aginis***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. - 06 Jul '06 02:27

more later 5 AM must sleeeeeeep*Originally posted by TommyC***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. - 06 Jul '06 07:21 / 2 editsMy 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. - 06 Jul '06 07:56

ok first*Originally posted by FabianFnas***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.