Notes

let the last 4 digits of n be abcd
then abcd mod 10000 = abcd^2 mod 10000 (we can ignore any digits greater than the 5th bc its greater than 10000 and thus isn't included in the mod calculation)
thus, using multiplication, we have that
d^2 + 10(cd + cd) + 100(bd+c^2+bd) + 1000(ad+bc+bc+ad) mod 10000 = abcd

note that the units digit of d^2 must be d
thus we have 2 cases
d = 1
d = 6

case 1: d = 1
1 + 10(2c) + 100(bc + c^2 + b) + 1000(a+bc+bc+a) mod 10000 = abc1
again, we can note here that the units digit of 2c must = c
we have one case, c = 0
we now have that
1 + 100(2b) + 1000(2a) mod 10000 = ab01
then we have that the units digit of 2b must = b
which is the same as c so b = 0
finally, we have that the units digit of 2a must = a
which is the same as c so a = 0
so we have 0001

case 2: d = 6
6 + 10(12c + 3) + 100(6b+c^2+6b) + 1000(6a + bc +bc +6a) mod 10000 = abc6
we note that the units digit of 12c+3 = c
we have one case, c = 7
we now have that
6 + 10(7) + 100(6b + 49 + 6b + 8) + 1000(6a + 14b + 6a) mod 10000 = ab76
we note that the units digit of 12b + 57 = b
we have one case, b = 3
we now have that
6 + 10(7) + 100(1) + 1000(12a + 42 + 9) mod 10000 = a376
we note that the units digit of 12a+51 = a
we have one case, a = 9
so we have 9376