Given the equation x^2-3x+1=0 prove that the sum of the squares of the roots is a natural number.

If x1 and x2 are the roots of the given equation,
substituted into equation, they verify it.

x1^2 - 3x1 + 1 =
0 (1)

x2^2 - 3x2 + 1 = 0
(2)

We'll add (1) + (2):

x1^2
+ x2^2 - 3(x1 + x2) + 2 = 0

Since we'll have to determine
the sum of the squares, we'll move to the right side of the equal sign, the rest of the
expression.

x1^2 + x2^2 = 3(x1 + x2) - 2
(*)

We'll apply Viete's relations for finding out x1 +
x2:

x1 + x2 = -b/a

x1 + x2 =
3

We'll substitute the sum of the roots into
(*):

x1^2 + x2^2 = 3*3 -
2

x1^2 + x2^2 = 9 -
2

x1^2 + x2^2 =
7

It is obvious that 7 is a
natural number, so the sum of the squares of the roots of the equation is a natural
number.