Conditions for $4^x+2^x+1$ to be a prime number.

For this proof, I will use the lemma that $b^n-1$ is always divisible by $b-1$ for $n\geq1$ . This is often proven by mathematical induction, but we can also consider the expansion of the dividend: $b^n-1=(b-1)(1+b+b^n+...+b^{n-1})$ . Rearranging the question: $\begin{aligned}4^x+2^x+1&=\left(2^x\right)^2+2^x+1\\ &=\frac{\left(2^x\right)^3-1}{2^x-1}\\ &=\frac{\left(2^3\right)^x-1}{2^x-1}\end{aligned}$ From the lemma, we can see that the numerator of this expression is always divisible by $2^3-1=7$ . But it is not true that the entire expression is always divisible by 7: if we let x equal 3, we get a prime number: $4^3+2^3+1=73\text{ is prime}$ The issue is the denominator: if the denominator is also a multiple of 7, then we have already once divided the numerator by 7, and we can no longer be certain if the whole expression is still divisible by 7. It turns out the denominator is divisible by 7 if x equals 3.

But there are other values of x that work: let $x=3^n k$ , where n is a non-negative integer and k is a positive integer. $\frac{\left(2^3\right)^{3^n k}-1}{2^{3^n k}-1}=\frac{\left(2^{3^{n+1}}\right)^k-1}{\left(2^{3^n}\right)^k-1}$ Now we can see that the numerator is always divisible by $2^{3^n+1}-1$ . The only way for the entire expression not to be divisible by $2^{3^n+1}-1$ is if the denominator is also divisible, which occurs when k=3: $\text{when }k=3:\,\frac{\left(2^{3^{n+1}}\right)^k-1}{\left(2^{3^n}\right)^k-1}=\frac{\left(2^{3^{n+1}}\right)^3-1}{\left(2^{3^n+1}\right)-1}$ If k=3 then $x=3^n k$ is a power of 3. This proves that the only possible values of x such that the expression $4^x+2^x+1$ could potentially be a prime number are powers of 3, and so we can conclude that $\boxed{x\text{ is indivisible by any prime number except 3}}$