Ans C

Ans C

The long way :

x² + y² + z² = 2(xy + 1)
x² + y² + z² = 2xy + 2
x² - 2xy + y² = 2 - z²
(x - y)² = 2 - z²
z² = 2 - (x - y)² ①

Next,

x + y + z = 2022
z = 2022 - (x + y)
z² = (2022 - (x + y))²
z² = 2022² - 2(2022)(x + y) + (x + y)² ②

① = ②

2 - (x - y)² = 2022² - 2(2022)(x + y) + (x + y)²

2 = (x - y)² + (x + y)² + 2022² - 2(2022)x - 2(2022)y

2 = 2x² + 2y² + 2022² - 2(2022)x - 2(2022)y

2 = 2x² - 4(1011)x + 2(1011)² + 2y² - 4(1011)y + 2(1011)²

[ Because 2022² = (2×1011)² = 2² × 1011² = 4 × 1011² and this can be split into 2(1011)² + 2(1011)²

Also, 2(2022) = 2(2 × 1011) = 4(1011) ]


Divide throughout by 2,

1 = x² - 2(1011)x + 1011² + y² - 2(1011)y + 1011²

1 = (x - 1011)² + (y - 1011)²

This actually fits the equation of a circle (x - a)² + (y - b)² = r², whereby the circle has radius 1 unit and centre (1011,1011)

From this, we can identify four coordinates of the circle with positive integers by simply moving vertically or horizontally from the centre by 1 unit to reach the circumference :

(1011,1010)
(1011,1012)
(1010,1011)
(1012,1011)

Then, sub these coordinates back into the equation x + y + z = 2022, to find z.

For (1011,1010) :

1011 + 1010 + z = 2022
2021 + z = 2022
z = 1

For (1011,1012) :

1011 + 1012 + z = 2022
2023 + z = 2022
z = -1
(does not fit the information that z is a positive integer)

For (1010,1011) :

1010 + 1011 + z = 2022
2021 + z = 2022
z = 1

For (1012, 1011) :

1012 + 1011 + z = 2022
2023 + z = 2022
z = -1
(does not fit the information that z is a positive integer)


Therefore, only (1010,1011) and (1011,1010) fit the requirements.

The two distinct solutions of x are 1010 and 1011

x1 + x2 = 1010 + 1011 = 2021

Answer is (C)

The short way :

x² + y² + z² = 2(xy + 1)
x² + y² + z² = 2xy + 2
x² - 2xy + y² = 2 - z²
(x - y)² = 2 - z²


Now x,y,z are all positive integers, and we know (x - y)² ≥ 0 for all real x and y values.

(x - y)² = 0 if x = y
(x - y)² > 0 if x ≠ y


So 2 - z² ≥ 0

The only possible positive integer value that fits this is z = 1, such that 2 - 1² = 1 > 0

Any other positive integer values of z will yield negative values of 2 - z².

This also tells us that (x - y)² = 1. So, x ≠ y

Then,

x - y = 1 or x - y = -1
x = y + 1 ① or x = y - 1 ②


Since x + y + z = 2022, and z = 1,

x + y + 1 = 2022
y = 2021 - x ③

Sub ③ into ①,

x = 2021 - x + 1
2x = 2022
x = 1011

Sub ③ into ②,

x = 2021 - x - 1
2x = 2020
x = 1010

So x1 + x2 = 1010 + 1011 = 2021

Answer is (C)