Equation:
x2+y2+z2=3x,y, and z are integers.
Formula used:
The algebraic identity:
(x+y+z)2=x2+y2+z2+2(xy+yz+zx)Calculations:
We are given
x2+y2+z2=3, where
x,y, and
z are integers.
Since
x,y, and
z are integers, their squares
(x2,y2,z2) must be non-negative integers
(0,1,4,9,...).
The only way to sum three non-negative integer squares to get 3 is if each square is 1
⇒x2=1,y2=1,z2=1This implies that
x,y, and z can only be either 1 or -1 .
Now, let
P=(xy+yz+zx) and
S=(x+y+z).
Using the algebraic identity:
(x+y+z)2=x2+y2+z2+2(xy+yz+zx)⇒S2=3+2PRearrange the equation to solve for P :
⇒2P=S2−3⇒P=()Now, we will consider all possible combinations of
x,y, and
z from
{1,−1} and calculate
S and
P for each case:
Case 1: All three integers are 1.
(x,y,z)=(1,1,1)S=1+1+1=3P=()=()=()=3 Case 2: Two integers are 1 , and one is -1 .
Possible combinations:
(1,1,−1),(1,−1,1),(−1,1,1)For any of these combinations,
S=1+1−1=1P=()=()=()=−1Case 3: One integer is 1, and two are -1.
Possible combinations:
(1,−1,−1),(−1,1,−1),(−1,−1,1)For any of these combinations,
S=1−1−1=−1P=()=()=()=−1Case 4: All three integers are -1.
(x,y,z)=(−1,−1,−1)S=−1−1−1=−3P=()=()=()=3The distinct values that
(xy+yz+zx) can have are 3 and -1 .
