"Almost same" square number?

Let the "Almost same" square number be ${ x }^{ 2 }$ and $n$ be the first $3$ digits of the number. Then,

${ x }^{ 2 }=1000n+\left( n+1 \right) =1001n+1\rightarrow { x }^{ 2 }-1\equiv 0\pmod {1001}$

Because of $1001=7\times11\times13$ , the equation can be changed into

$\begin{cases} \left( x-1 \right) \left( x+1 \right) \equiv 1\pmod {7} \\ \left( x-1 \right) \left( x+1 \right) \equiv 1\pmod {11} \\ \left( x-1 \right) \left( x+1 \right) \equiv 1\pmod {13} \end{cases}\rightarrow \begin{cases} x\equiv \pm 1\pmod {7} \\ x\equiv \pm 1\pmod {11} \\ x\equiv \pm 1\pmod {13} \end{cases}$

After consider different cases You will get these answers: $x\equiv 1, 155, 274, 428, 573, 727, 846, 1000\pmod {1001}$ . However, by definition,

$100000<{ x }^{ 2 }<1000000\rightarrow 316<x<1000$

Therefore, $x=428, 573, 727, 846$ and the answer is ${ 428 }^{ 2 }+{ 573 }^{ 2 }+{ 727 }^{ 2 }+{ 846 }^{ 2 }=183184+328329+528529+715716=\boxed {1755758}$

Prints all $4$ cases then prints their sum:

1
2
3
4
5
6
7
8
9

import math
a = 0
for i in range(100, 999):
    z = 1001*i + 1
    y = math.sqrt(z)
    if ( y - round(y) == 0):
        print(z)
        a += z
print("sum = " + str(a))

Total[Table[ If[Sqrt[1000 n + n + 1] [Element] Integers, 1000 n + n + 1, Nothing], {n, 100, 998}]] gives 1755758.

The numbers are 183184, 328329, 528529 and 715716.