Question

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.

Answer 1

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