Geodesic
A note on factorization of the Fermat numbers and their factors of the form $3h2\sp n+1$
Mathematica Bohemica, Tome 119 (1994) no. 4, pp. 437-445

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We show that any factorization of any composite Fermat number $F_m={2^{2}}^m+1$ into two nontrivial factors can be expressed in the form $F_m=(k2^n+1)(\ell2^n+1)$ for some odd $k$ and $\ell, k\geq 3, \ell \geq 3$, and integer $n\geq m+2, 3n2^m$. We prove that the greatest common divisor of $k$ and $\ell$ is 1, $k+\ell\equiv 0\ mod 2^n,\ max(k,\ell)\geq F_{m-2}$, and either $3|k$ or $3|\ell$, i.e., $3h2^{m+2}+1|F_m$ for an integer $h\geq 1$. Factorizations of $F_m$ into more than two factors are investigated as well. In particular, we prove that if $F_m=(k2^n+1)^2(\ell2^j+1)$ then $j=n+1,3|\ell$ and $5|\ell$.
We show that any factorization of any composite Fermat number $F_m={2^{2}}^m+1$ into two nontrivial factors can be expressed in the form $F_m=(k2^n+1)(\ell2^n+1)$ for some odd $k$ and $\ell, k\geq 3, \ell \geq 3$, and integer $n\geq m+2, 3n2^m$. We prove that the greatest common divisor of $k$ and $\ell$ is 1, $k+\ell\equiv 0\ mod 2^n,\ max(k,\ell)\geq F_{m-2}$, and either $3|k$ or $3|\ell$, i.e., $3h2^{m+2}+1|F_m$ for an integer $h\geq 1$. Factorizations of $F_m$ into more than two factors are investigated as well. In particular, we prove that if $F_m=(k2^n+1)^2(\ell2^j+1)$ then $j=n+1,3|\ell$ and $5|\ell$.
DOI : 10.21136/MB.1994.126115
Classification : 11A51, 11Y05
Keywords: congruence properties; Fermat numbers; prime numbers; factorization; squarefreensess 
Křížek, Michal; Chleboun, Jan. A note on factorization of the Fermat numbers and their factors of the form $3h2\sp n+1$. Mathematica Bohemica, Tome 119 (1994) no. 4, pp. 437-445. doi: 10.21136/MB.1994.126115
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