if the last three digits of the square of a positive integer are the same and nonzero,we say that the positive integer is good. for example,?

The last 3 digits of the number must be 1000n + 111 , 1000n + 444 , 1000n + 555 , 1000n + 666
or 1000n + 999.
But 1000n + 111 (mod 8) ≡ 7 , 1000n + 555 (mod 8) ≡ 3 , 1000n + 666 (mod 4) ≡ 2 , 1000n + 999 (mod 8) ≡ 7 are NOT perfect square since odd square (mod 8) ≡ 1 and even square (mod 4) ≡ 0.

Let (10x + 2)² = 100n + 44
100x² + 40x + 4 = 100n + 44
5x² + 2x = 5n + 2
5(x² - n) = 2(1 - x)
1 - x (mod 5) ≡ 0 ⇒ x = n = 1 or x = 6 , n = 38 , we have 12² = 144 and 62² = 3844.

Let (10x + 8)² = 100n + 44
100x² + 160x + 64 = 100n + 44
5x² + 8x + 1 = 5n
8x + 1 = 5(n - x² )
8x + 1 (mod 5) ≡ 0 ⇒ x = 3 , n = 14 or x = 8 , n = 77 , we have 38² = 1444 and 88² = 7744.

So the second number at least 3 digits.

Let (100a + 12)² = 1000x + 444
10000a² + 2400a + 144 = 1000x + 444
100a² + 24a - 3 = 10x
10(10a² - x) = 3 - 24a , no solution for a.

Let (100a + 38)² = 1000x + 444
10000a² + 7600a + 1444 = 1000x + 444
100a² + 76a + 10 = 10x
76a = 10(x - 10a² - 1)
19a = 5(x - 10a² - 1)
⇒ a = 5 , x = 289 , we have 538² = 289444

Let (100a + 62)² = 1000x + 444
10000a² + 12400a + 3844 = 1000x + 444
100a² + 124a + 34 = 10x
124a + 34 = 10(x - 10a²)
a = 4 , x = 213 , we have 462² = 213444.
No need to consider a = 9 > 4.

Let (100a + 88)² = 1000x + 444
10000a² + 17600a + 7744 = 1000x + 444
100a² + 176a + 73 = 10x
176a + 73 = 10(x - 10a²) , no solution for a.

All in all 462 is the second number.