Let the roots be a,b,c,d,e.
Then (x-a)(x-b)(x-c)(x-d)(x-e) = 0
Expanding,
x^5
- x^4 (a+b+c+d+e)
+x^3 (ab + ac + ... cd +ce + de) [sum of 10 pairwise products]
- x^2 (abc + abd + ... + bce + bde + cde) [sum of 10 three way products]
+ x (bcde + acde + abde + abce + abcd)
- (abcde)= 0
But x^5 - x^2 +1 = 0
so (a+b+c+d+e) = 0
(ab + ac + ... cd +ce + de) = 0 [sum of 10 pairwise products]
(abc + abd + ... + bce + bde + cde) = 1 [sum of 10 three way products]
(bcde + acde + abde + abce + abcd) = 0
(abcde) = -1
I expect it is possible to use these identities to assign a value to
(a^2+1)(b^2+1)(c^2+1)(d^2+1)(e^2+1)
Algebraic approach
