We will call this positive integer x.

We will call the other integer y.

x=2y-4

x^{2}+y^{2}=772

We are looking for only integer solutions to the following equations.

First, plug in the first equation into the second,

(2y-4)^{2}+y^{2}=772

4y^{2}-16y+16+y^{2}=772

5y^{2}-16y+16=772

5y^{2}-16y-756=0

Use the quadratic formula:

y = 16±√(256-4(5)(-756))/(10)

y=(16±124)/(10)

We only want positive integers,

y=(16+124)/10

y=14

Plug back into the first equation,

x=2(14)-4

x=28-4

x=24.

Therefore, the two positive integer solutions to this system of equations are x=24 and y=14.