The sum of the squares of two positive integers is 394. If one integer is 2 less than the other and the larger integer is x. Find the integers.

Larger is x

the other is called y

sum of squares is 394

x^2+y^2=394

one is 2 less that other

x>y so

y+2=x

subsitute y+2 for x

(y+2) ^2+y^2=394

expand

y^2+4y+4+y^2=394

add like terms

2y^2+4y+4=394

divideboth sides by 2

y^2+2y+2=197

subtract 197 from both sides

y^2+2y-195=0

factor

(y+15) (y-13) = 0

set each to zero

y+15=0

y=-15

impossible since it is stated that they are positive so get rid of this solution

y-13=0

y=13

find other

y+2=x

13+2=x

15=x

numbers are 13 and 15