repeats

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.

Maybe I was a little vague here and there. Good that you are abservant.
It was crystal clear when I formulated the 'proof'.

You wrote:
"437^2 = 190969 = 3*63656+1~1mod3
finally see an example above for a number which when repeated is a square."

Please, again, show me the number, R,when repeated, RR, is a square.

Originally posted by aginis
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
5882352941176470588235294117647058823529411764705882352941176470588235294117648

aginis is correct here, well done!

But there's a smaller example than the one given above.

It all depends on whether or not we can find a square factor of 10^n + 1 for some n.

In fact

10^11 + 1 = 11^2 x 23 x 4093 x 8779

and so (10^11 + 1) x 23 x 4093 x 8779 is square. Unfortunately this square is not the repeat of 23 x 4093 x 8779, which has too few digits, but we can simply multiply it by 4^2 to get it in the right range.

In fact 23 x 4093 x 8779 x 4^2 = 13223140496 and

1322314049613223140496 = 36363636364^2

where of course 36363636364 = 11 x 23 x 4093 x 8779 x 4.