if x and y are positive integers such that x^2+y^2=41, then what is the value of (x+y)^2?

Question

if x and y are positive integers such that x^2+y^2=41, then what is the value of (x+y)^2?

how did you get that?

like how did you get that without guessing and testing...?

Answer 1

The only two solutions to the first equation are x=4,y=5 and x=5,y=4. Both yield (x+y=9) and 9^2=81.

Answer 2

As x and y are positive integers, there is a limit of options for x and y. Both must be lower then 7. So the correct math solution may be found by verification of each option:
x= 1,2,3,4,.5,6
y= 1,2,3,4,5,6 and here are only two answers: x=5 and y = 4 or x=4 and y = 5

In each case (x+y)^2 = (9)^2 = 81

Answer 3

(x+y)^2 = 41+2xy
x=5 and y = 4 satisfy the equation x^2 + y^2 = 41
41+2xy = 41+ 40 = 81 = (5+4)^2