Question

factorize x^2+y^2+z^2/4+2xy-yz-zx

Answer 1

Take the differences by pairs:

(z – y)(x + y + z) = 20 = 2*2*5

(z – x)(x + y + z) = 30 = 3*2*5

(y – x)(x + y + z) = 10 = 1*2*5

Clearly, x, y & z are distinct.

To find a solution in positive integers:

There is no way to partition 5 into three distinct integers so x + y + z = 10
Substituting:
y = x + 1
z = x + 3

x + x + 1 + x + 3 = 10
3x = 6
x = 2
y = 3
z = 5

This is the easy solution.