Find the unique degree $6$ polynomial

Question

What makes it unique is the following requirements on the polynomial:

1. It's coefficients are either $0$ or $1$
2. It has no rational roots
3. The absolute value of $p(x)$ isn't prime for every integer.

Just for this problem, $1$ isn't a prime.

Answer 1

Here it is: $$f(x)=x^6+x^4+x^2+1=(x^2+1)(x^4+1)$$

It's also even, so $f(x)=f(-x)$. We get $f(0)=1$, and $f(n)$ has two factors greater than $1$ for $n \geq 1$ due to the polynomial's factorization.

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To prove it's unique:

• The coefficient of $x^6$ is $1$, otherwise it doesn't have degree $6$. The constant term is $1$, otherwise it has root $0$.
• It must have $4$ non-zero coefficients (otherwise $f(1) \in \{2,3,5\}$, and $x^6+x^5+x^4+x^3+x^2+x+1$ evaluated at $x=1$ gives $7$ (which is prime)).
• What's left has $f(-1) \in \{0,2\}$ (giving either a root or a prime) except for the above polynomial.