Find a square number have 6 numeral so as to : first three numberal are the same last three numeral?

Question

Ex:square number are:

1; 2; 4 ;9 ;14 ;25; 36; 49; 64; 81...

Answer 1

There is no such number. Here is the proof:
We want to solve
x² = 10^5 a + 10^4 b + 10^3 c + 100a + 10 b + c.
and we may assume that x is not 0.
This yields
x² = 1000(100a + 10b + c) + 100a + 10b + c
= 1001(100a + 10b + c).
But 1001 = 7*11*13 is square free, so that
forces 100a +10b+c to be divisible by 1001.
This is impossible, since 100a + 10b + c is only
a 3 digit number

Answer 2

abcabc=X^2
a (10) ^ 5 + b(10) ^ 4 + c(10) ^ 3 +a(10) ^ 3 + b(10) ^ 2 +c=

= (10 ^ 3 + 1) [a(10) ^ 2 +b(10) +c] = 11*91*[a(10) ^ 2 +b(10) +c]
aneweris is no sach number becouse ....???