Let ABCD is a square.

Let the coordinate of vertices A and C are (3, 4) and (1, -1) respectively.

Let the coordinate of vertex B be (x, y)

Since, in square, length of all sides are equal

=> AB = BC

=> √{(x - 3)² + (y - 4)² } = √{(x - 1)² + (y + 1)² }

Squaring on both side, we get

=> (x - 3)² + (y - 4)² = (x - 1)² + (y + 1)²

=> x² - 6x + 9 + y² - 8y + 16 = x² - 2x + 1 + y² + 2y + 1

=> x² - 2x + 1 + y² + 2y + 1 - x² + 6x - 9 - y² + 8y - 16 = 0

=> 4x + 10y - 23 = 0

=> 4x + 10y = 23

Now in right angle triangle ABC,

      AB² + BC² = CA²

=> (x - 3)² + (y - 4)² + (x - 1)² + (y + 1)² = (3 - 1)² + (4 + 1)²

=> x² - 6x + 9 + y² - 8y + 16 + x² - 2x + 1 + y² + 2y + 1 = 4 + 25

=> 2x² + 2y² - 8x - 6y - 2 = 0 

=> x² + y² - 4x - 3y - 1 = 0

=> {(23 - 10y)/4}² + y² - 4{(23 - 10y)/4} - 3y - 1 = 0

=> 529 + 100y² - 460y + 16y² - 368 + 160y - 48y - 16 = 0

=> 116y² - 348y + 145 = 0

=> 29(4y² - 12y + 5) = 0

=> 4y² - 12y + 5 = 0

=> 4y² - 10y - 2y + 5 = 0

=> 2y(2y - 5) - 1(2y - 5) = 0

=> (2y - 5) * (2y - 1) = 0

=> y = 1/2, 5/2

When y = 1/2, we get

x = (23 - 10 * 1/2)/4 = (23 - 5)/4 = 18/4 = 9/2

When y = 5/2, we get

x = (23 - 10 * 5/2)/4 = (23 - 25)/4 = -2/4 = -1/2

So, the other vertices of the square are (9/2, 1/2) and (-1/2, 5/2)