Divisibility of $6^{2^n}+ 8^{2^n} +12^{2^n}+14^{2^n}+16^{2^n}+18^{2^n} +24^{2^n} +28^{2^n}+42^{2^n}$

Prove or disprove that for all natural $n$

$$6^{2^n}+ 8^{2^n} +12^{2^n}+14^{2^n}+16^{2^n}+18^{2^n} +24^{2^n} +28^{2^n}+42^{2^n}$$

is divisible by $259$.

I tried to apply mathematical induction, but the step is overly cumbersome.

Some thoughts. Since $2^{2^{n+1}}=(2^{2^n})^2$‚ it is fairly easy to get:

$\begin{array}{c|ccccc} \mod & 7 & 37 \\ \hline 6^2 &1 &36 \\ 6^4 &1 &1 \\ 6^8 &1 &1 \\ 6^{16} &1 &1 \\ 6^{32} &1 &1 \\ 6^{64} &1 &1 \\ 6^{128} &1 &1 \end{array}$ $\begin{array}{c|ccccc} \mod & 7 & 37 \\ \hline 8^2 &1 &27 \\ 8^4 &1 &26 \\ 8^8 &1 &10 \\ 8^{16} &1 &26 \\ 8^{32} &1 &10 \\ 8^{64} &1 &26 \\ 8^{128} &1 &10 \end{array}$ $\begin{array}{c|ccccc} \mod & 7 & 37 \\ \hline 12^2 &4 &33 \\ 12^4 &2 &16 \\ 12^8 &4 &34 \\ 12^{16} &2 &9 \\ 12^{32} &4 &7 \\ 12^{64} &2 &12 \\ 12^{128} &4 &33 \end{array}$

$\begin{array}{c|ccccc} \mod & 7 & 37 \\ \hline 14^2 &0 &11 \\ 14^4 &0 &10 \\ 14^8 &0 &26 \\ 14^{16} &0 &10 \\ 14^{32} &0 &26 \\ 14^{64} &0 &10 \\ 14^{128} &0 &26 \end{array}$ $\begin{array}{c|ccccc} \mod & 7 & 37 \\ \hline 16^2 &4 &34 \\ 16^4 &2 &9 \\ 16^8 &4 &7 \\ 16^{16} &2 &12 \\ 16^{32} &4 &33 \\ 16^{64} &2 &16 \\ 16^{128} &4 &34 \end{array}$ $\begin{array}{c|ccccc} \mod & 7 & 37 \\ \hline 18^2 &2 &28 \\ 18^4 &4 &7 \\ 18^8 &2 &12 \\ 18^{16} &4 &33 \\ 18^{32} &2 &16 \\ 18^{64} &4 &34 \\ 18^{128} &2 &9 \end{array}$

$\begin{array}{c|ccccc} \mod & 7 & 37 \\ \hline 24^2 &2 &21 \\ 24^4 &4 &34 \\ 24^8 &2 &9 \\ 24^{16} &4 &7 \\ 24^{32} &2 &12 \\ 24^{64} &4 &33 \\ 24^{128} &2 &21 \end{array}$ $\begin{array}{c|ccccc} \mod & 7 & 37 \\ \hline 28^2 &0 &7 \\ 28^4 &0 &12 \\ 28^8 &0 &33 \\ 28^{16} &0 &16 \\ 28^{32} &0 &34 \\ 28^{64} &0 &9 \\ 28^{128} &0 &7 \end{array}$ $\begin{array}{c|ccccc} \mod & 7 & 37 \\ \hline 42^2 &0 &25 \\ 42^4 &0 &33 \\ 42^8 &0 &16 \\ 42^{16} &0 &34 \\ 42^{32} &0 &9 \\ 42^{64} &0 &7 \\ 42^{128} &0 &12 \end{array}$

Some Mathematica code:

Format[primeFactorsFormat[n_Integer]] := 
 CenterDot @@ Superscript @@@ FactorInteger[n] //. _[x_] | _[x_, 1] :> x
tab = Table[{n, primeFactorsFormat[
     6^2^n + 8^2^n + 12^2^n + 14^2^n + 16^2^n +
      18^2^n + 24^2^n + 28^2^n + 42^2^n]}, {n, 1, 5}];
TableForm[tab, TableHeadings -> {None, {"n", "factorization"}}]

Solution 1:

By FLT: $a^{2^n}\equiv a^{2^n\bmod {(p-1)}}\pmod {p}$ when $p\nmid a$. Also $259=7\cdot 37$.

The statement is false when $n=0$. I assume $n\neq 0$.

$2^n\equiv \{2,4\}\pmod {6},\: 2^n\equiv \{2,4,8,16,32,28,20\}\pmod {36}$

Now just check that $6^m+ 8^m +12^m+16^m+18^m +24^m$ is divisible by $7$ when $m\in\{2,4\}$ (simple to check)

and that $6^m+ 8^m +12^m+14^m+16^m+18^m +24^m +28^m+42^m$ is divisible by $37$ when $m\in\{2,4,8,16,32,28,20\}$ (lengthy, but all it requires are elementary modular arithmetic manipulations. I used WolframAlpha (several times, changing the value of $m$)).

Indeed all of this holds, so $259$ divides the expression $\forall n\ge 1$.

Solution 2:

$6^{2^n}+ 8^{2^n} +12^{2^n}+14^{2^n}+16^{2^n}+18^{2^n} +24^{2^n} +28^{2^n}+42^{2^n}=$

$2^{2^n}(2^{2^n}+3^{2^n}+1)(2^{2^{n+1}}+3^{2^n}+7^{2^n})$

Now simply prove by induction:

• $\forall{n}\in\mathbb{N}:7|(2^{2^n}+3^{2^n}+1)$
• $\forall{n}\in\mathbb{N}:37|(2^{2^{n+1}}+3^{2^n}+7^{2^n})$

Solution 3:

$$6^{2^n}+ 8^{2^n} +12^{2^n}+14^{2^n}+16^{2^n}+18^{2^n}+24^{2^n} +28^{2^n}+42^{2^n} =$$ $$= 2^{2^n}(3^{2^n}+ 4^{2^n} +6^{2^n}+7^{2^n}+8^{2^n}+9^{2^n}+12^{2^n} +14^{2^n}+21^{2^n}) =$$ $$= 2^{2^n}((3^{2^n} +6^{2^n}+9^{2^n})+ (4^{2^n}+8^{2^n}+12^{2^n})+(7^{2^n} +14^{2^n}+21^{2^n})) =$$ $$= 2^{2^n}(1 +2^{2^n}+3^{2^n})(3^{2^n}+4^{2^n}+7^{2^n})$$

$\begin{array}{c|ccccc} \mod & 7 \\ \hline 2^2 &4 \\ 2^4 &2 \\ 2^8 &4 \end{array}$ $\begin{array}{c|ccccc} \mod & 7 \\ \hline 3^2 &2 \\ 3^4 &4 \\ 3^8 &2 \end{array}$

$\begin{array}{c|ccccc} \mod & 37 \\ \hline 3^2 &9 \\ 3^4 &7 \\ 3^8 &12 \\ 3^{16} &33 \\ 3^{32} &16 \\ 3^{64} &34 \\ 3^{128} &9 \end{array}$ $\begin{array}{c|ccccc} \mod & 37 \\ \hline 4^2 &16 \\ 4^4 &34 \\ 4^8 &9 \\ 4^{16} &7 \\ 4^{32} &12 \\ 4^{64} &33 \\ 4^{128} &16 \end{array}$ $\begin{array}{c|ccccc} \mod & 37 \\ \hline 7^2 &12 \\ 7^4 &33 \\ 7^8 &16 \\ 7^{16} &34 \\ 7^{32} &9 \\ 7^{64} &7 \\ 7^{128} &12 \end{array}$

etc.