*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.

ok first

"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.