The positive integers are 21 and 38, where one integer is 4 less than twice the other. The sum of their squares is 772, satisfying the given conditions.
Let's denote the two positive integers as x and y, with x > y. According to the given information:
1. x is 4 less than twice y: x = 2y - 4.
2. The sum of their squares is 772:

Now, substitute the expression for x from the first equation into the second:
![\[(2y - 4)^2 + y^2 = 772\]](https://img.qammunity.org/2024/formulas/mathematics/college/5x7o9cf55k96eme2ge90qzngs20envlkpm.png)
Expand and simplify:
![\[4y^2 - 16y + 16 + y^2 = 772\]](https://img.qammunity.org/2024/formulas/mathematics/college/ydqoc7cy8hq5vkwl1mvsfxr08mqnfgp81f.png)
Combine like terms:
![\[5y^2 - 16y + 16 = 772\]](https://img.qammunity.org/2024/formulas/mathematics/college/uriunfzjguygidhfl6j3jabyh7fzqxk1ek.png)
Subtract 772 from both sides:
![\[5y^2 - 16y - 756 = 0\]](https://img.qammunity.org/2024/formulas/mathematics/college/m5z4zye2o4yijk4hlewgagyt8v5mm6gbrm.png)
Factor the quadratic equation:
(5y + 36)(y - 21) = 0
This gives two possible solutions for y:
(extraneous) or y = 21.
Substitute y = 21 back into the first equation to find x:
x = 2(21) - 4 = 38
Therefore, the two positive integers are 21 and 38.