Is this a joke?

Continuing the proof from the sister problem ...

Note that $7^2 \equiv 1 \pmod{16}$ , which means that $7^{2^n} \equiv 1 \pmod{2^{n+3}}$ for all integers $n \ge 1$ . Similarly $7^4 \equiv 1 \pmod{25}$ , and hence $7^{4 \times 5^n} \equiv 1 \pmod{5^{n+2}}$ for all $n \ge 0$ . We certainly deduce that $7^{10^n} \equiv 1 \pmod{10^{n+2}}$ for all $n\ge 2$ . We also note that $7^{20} \equiv 1 \pmod{1000}$ .

Letting $a_n = {}^{(n)}7$ denote that level $n$ tetration of $7$ , we see that $a_2 = 7^7 \equiv 3 \pmod{20}$ , and hence $a_3 \equiv 7^3 \equiv 343 \pmod{1000}$ . Thus $\begin{array}{rclcrcl} a_3 & \equiv & 43 \pmod{10^2} & \Rightarrow & a_4 & \equiv & 7^{43} \equiv 2343 \pmod{10^4} \\ a_4 & \equiv & 343 \pmod{10^3} & \Rightarrow & a_5 & \equiv & 7^{343} \equiv 72343 \pmod{10^5} \\ a_5 & \equiv & 2343 \pmod{10^4} & \Rightarrow & a_6 & \equiv & 7^{2343} \equiv 172343 \pmod{10^6} \\ a_6 & \equiv & 72343 \pmod{10^5} & \Rightarrow & a_7 & \equiv & 7^{72343} \equiv 5172343 \pmod{10^7} \\ a_7 & \equiv & 172343 \pmod{10^6} & \Rightarrow & a_8 & \equiv & 7^{172343} \equiv 65172343 \pmod{10^8} \end{array}$ Moreover $a_n \equiv 7^{172343} \pmod{10^8} \;\Rightarrow\; a_n \equiv 65172343 \pmod{10^8} \;\Rightarrow\; a_n \equiv 172343 \pmod{10^6} \;\Rightarrow\; a_{n+1} \equiv 7^{172343} \pmod{10^8}$ Thus we deduce that $a_n \equiv 7^{172343} \pmod{10^8}$ for all $n \ge 8$ . While we have used the fact that $7^{10^6} \equiv 1 \pmod{10^8}$ , the results in the first paragraph show that $7^{500000} \equiv 1 \pmod{10^8}$ . In fact $500000$ is the order of $7$ modulo $10^8$ , and so the smallest possible positive integer $n$ such that $a_{2018} \equiv 7^n \pmod{10^8}$ is $\boxed{172343}$ .

---

What is interesting is the developing pattern. Note that $a_3 \equiv a_2 \equiv 43 \pmod{10^2}$ . If $m \ge 3$ and $a_m \equiv a_2 \pmod{10^2}$ , then $a_{m+1} \equiv 7^{a_m} \equiv 7^{a_2} \equiv a_3 \equiv a_2 \pmod{10^4}$ , so certainly $a_m \equiv a_2 \pmod{10^2}$ . By induction, then, $a_m \equiv a_2 \pmod{10^2}$ .

Suppose now that $n \ge 2$ and that $a_m \equiv a_n \pmod{10^n}$ for all $m > n$ . Then $a_{m+1} \equiv 7^{a_m} \equiv 7^{a_n} \equiv a_{n+1} \pmod{10^{n+2}}$ , and so certainly $a_{m+1} \equiv a_{n+1} \pmod{10^{n+1}}$ for all $m > n$ , and hence $a_m \equiv a_{n+1} \pmod{10^{n+1}}$ for all $m > n+1$ . Thus we deduce that $a_m \equiv a_n \pmod{10^n}$ for all $m > n \ge 2$ .

The trick I used was to determine that $500,000$ is the smallest $n$ such that $7^n=1$ mod $10^8$ (a nice little fact). I determined that fact by writing the following (Matlab) code: t=7;counter=1; while t~=1, counter=counter+1;t=mod(7 t,10^8);end. The output of the counter gives the answer of $500,000$ . Then I wrote the following code to finish the job: a=7; for n=2:2018 a=mod(a,500000); m=1; for j=1:a m=mod(7 m,10^8); end a=m; end t=1; counter = 0; while t~=a counter = counter +1; t = mod(7*t,10^8); end. Again, the output of the counter gives the final answer of 172343.

I wrote a code to find 7 hyperexponentiated 2018 times, and then bruteforced till I found a solution. Very cheap method but it worked xP