Perfect squares

We have

$\displaystyle \begin{aligned} 1458 + x = y^2 \\ 3258 + x = z^2 \end{aligned}$

for positive integers $x, y, z$ .

Subtracting the first equation from the second gives us $(z + y)(z - y) = 1800$ . Letting $a = z + y$ and $b = z - y$ , we want to find $a, b$ such that $ab = 1800$ .

$a$ and $b$ must both be even, so we consider the even factors of 1800:

$(2, 900), (4, 450), (6, 300), (10, 180), (12, 150), (18, 100), (20, 90), (30, 60), (36, 50)$

giving us the following pairs of linear equations:

$\begin{aligned} z + y & = 2 & z - y & = 900 & \qquad (1) \\ z + y & = 900 & z - y & = 2 & \qquad (2) \\ z + y & = 4 & z - y & = 450 & \qquad (3) \\ z + y & = 450 & z - y & = 4 & \qquad (4) \\ z + y & = 6 & z - y & = 300 & \qquad (5) \\ z + y & = 300 & z - y & = 6 & \qquad (6) \\ z + y & = 10 & z - y & = 180 & \qquad (7) \\ z + y & = 180 & z - y & = 10 & \qquad (8) \\ z + y & = 12 & z - y & = 150 & \qquad (9) \\ z + y & = 150 & z - y & = 12 & \qquad (10) \\ z + y & = 18 & z - y & = 100 & \qquad (11) \\ z + y & = 100 & z - y & = 18 & \qquad (12) \\ z + y & = 20 & z - y & = 90 & \qquad (13) \\ z + y & = 90 & z - y & = 20 & \qquad (14) \\ z + y & = 30 & z - y & = 60 & \qquad (15) \\ z + y & = 60 & z - y & = 30 & \qquad (16) \\ z + y & = 36 & z - y & = 50 & \qquad (17) \\ z + y & = 50 & z - y & = 36 & \qquad (18) \end{aligned}$

The respective solutions to these systems of linear equations are

$\begin{aligned} y & = -449 & z & = 451 & \qquad (1)' \\ y & = 449 & z & = 451 & \qquad (2)' \\ y & = -223 & z & = 227 & \qquad (3)' \\ y & = 223 & z & = 227 & \qquad (4)' \\ y & = -147 & z & = 153 & \qquad (5)' \\ y & = 147 & z & = 153 & \qquad (6)' \\ y & = -85 & z & = 95 & \qquad (7)' \\ y & = 85 & z & = 95 & \qquad (8)' \\ y & = -69 & z & = 81 & \qquad (9)' \\ y & = 69 & z & = 81 & \qquad (10)' \\ y & = -41 & z & = 59 & \qquad (11)' \\ y & = 41 & z & = 59 & \qquad (12)' \\ y & = -35 & z & = 55 & \qquad (13)' \\ y & = 35 & z & = 55 & \qquad (14)' \\ y & = - 15 & z & = 45 & \qquad (15)' \\ y & = 15 & z & = 45 & \qquad (16)' \\ y & = - 7 & z & = 43 & \qquad (17)' \\ y & = 7 & z & = 43 & \qquad (18)' \end{aligned}$

Because $y > 0$ , we can eliminate half of these. Now, recall that $x = y^2 - 1458 > 0 \Longrightarrow y > 39$ .

Using $x = y^2 - 1458 = z^2 - 3258$ , we solve the remaining solutions $\left(y > 39, z\right)$ for $x$ :

$\begin{aligned} x & = 200143 & \qquad (2)' \\ x & = 48271 & \qquad (4)' \\ x & = 20151 & \qquad (6)' \\ x & = 5767 & \qquad (8)' \\ x & = 3303 & \qquad (10)' \\ x & = 223 & \qquad (12)' \end{aligned}$

Of those, only $223$ is a 3-digit integer. Our answer is $x = \boxed{223}$ .