Prove that the number $3^{3^n} + 1$ has at least $2n + 1$ prime factors.

$$\large(3^{2\times3^k} - 3^{3^k} + 1) = (3^{3^k}-3^{(3^k+1)/2}+1)(3^{3^k}+3^ {(3^k+1)/2}+1)$$


We have: $$3^{2\cdot3^k} - 3^{3^k} + 1=(3^{3^k}+3^{\frac{3^k+1}{2}}+1)(3^{3^k}-3^{\frac{3^k+1}{2}}+1)$$ And for $k>0$ both factors are greater than one. This factorization can be deduced from the fact that $f(2)=387400807=19441\cdot19927$ and that both factors are close to each other.

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Prime Numbers

Elementary Number Theory

Prime Factorization

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