Question

The sum of squares of three positive integers is 323. If the sum of squares of two numbers is twice the third, their product is

Answer 1

Let the three positive integers be x, y and z.

Sum of the squares of the three numbers = x² + y² + z²

According to question,
Sum of the squares of the three numbers=323
x² + y² + z²= 323.........(i)

The sum of the squares of the two numbers =x²+y²
According to question,
The sum of the squares of the two numbers =  Twice the third
The sum of the squares of the two numbers = 2z
x²+y²=2z.............(ii)

put x²+y²=2z frm(ii) to (i)
x² + y² + z²= 323
2z + z² = 323 

Subtract 323 from both sides:
2z + z² - 323 =0
Solve the quadratic equation by factoring:
z²+19z - 17z -323 =0
z(z+19)-17(z+19) = 0
(z+19)(z-17)=0

Applying zero product rule:
z+19=0 or z-17=0
z=-19 or z=17

Since, the integers are positive, we reject z=-19.

Therefore, the 3rd number is z=17.

put z= 17 in (ii)
x²+y²=2z
x²+y²=2(17)
x²+y²=34

Express 34 as a sum of two squares, 34=9+25(only possibility):
x²+y²=9+25
x²+y²=3²+5²

Then, the other two numbers are x=3 and y=5.

Threfore, the product of the three numbers =x*y*z = 3*5*17=255


Answer: 255