let the two positive integers be 'x' and 'y'

two positive integers differ by 3

y = x + 3

the sum of their squares is 117

x^2 + y^2 = 117

plug y = x + 3 into the last equation

x^2 + (x + 3)^2 = 117

by solving we find 2 roots and consider the positive

x = 6

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for x = 6 we have y = 6+3 = 9

the two numbers are 6 and 9.