Let f(n)= n^10+n^5+1
Denote the two non-real cube roots of unity as w and w^2
Then, f(w)=w^10+w^5 +1= w+w^2+1=0
Also, f(w^2)=w^20+w^10+1=w^2+w+1=0
Accordingly, f(n) is divisible by :
(n- w )(n- w^2)=n^2-(w^2+w )+w^3=n^2+n:+1
Then, dividing f(n) by n^2+n+1, we have a factorization of f(n) as:
f(n)=(n^2+n+1)(n^8-n^7+n^6-n^5+n^4-n^3+n^2-n+1)
=> f(n)=(n^2+n+1)*g(n), where:
g( n)=(n -1)(n^7+n^5+n^3+n)+1))
For n>=2, we observe that: n^2+n+1>=7, and:
g(n)>1*(128+32+8+2)+1=171
Threfore, their product f(n) can NEVER BE PRIME.
This is a contradiction.
Now, f(0)=1, which is NOT a prime number.
f(1)=1+1+1=3, which IS A PRIME NUMBER.
Consequently n=1, provides the solitary value amongst all natural numbers, such that the given expression is a prime number..
Edited on December 31, 2021, 11:37 am
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