Prime factors of a sum of special power towers

Question

Denote

$$x(n) = 2 \uparrow 3 \uparrow ... \uparrow n$$

and

$$y(n) = n \uparrow (n-1) \uparrow ... \uparrow 3 \uparrow 2$$

Finally denote

$$z(n) = x(n) + y(n)$$

So, the first few numbers z(n) are

$$z(2) = 2 + 2 = 4$$ $$z(3) = 2^3 + 3^2 = 17$$ $$z(4) = 2^{3^4}+4^{3^2} = 2^{81} + 4^9$$ $$z(5) = 2^{3^{4^5}} + 5^{4^{3^2}} = 2^{3^{1024}} + 5^{2^{18}}$$

For odd $n\ge3$, every prime factor p > 2 of z(n) has -2 as a quadratic residue because z(n) is of the form $x^2+2y^2$ with gcd(x,y)=1. For z(5), I checked the prime factors up to $10^7$ and only found 3,11,59 and 281.

The first number z(n) without very small prime factors is z(9).

Could anyone search for primefactors of z(9) upto, lets say, $10^7$ ?

Answer 1

My quick computer search suggests the smallest factor of $z(9)$ should be $523$. Confirmation (of it being a factor; not necessarily the smallest one):

First addend:

• $2^{3\uparrow 4\uparrow 5\ldots\uparrow 9}\equiv 2^{(3\uparrow 4\uparrow 5\ldots\uparrow 9)\ \bmod\ 522}\pmod {523}$
  • $522=2\times 3^2\times 29$
  • $3^{4\uparrow 5\ldots\uparrow 9}\equiv 1\ \pmod 2$
  • $3^{4\uparrow 5\ldots\uparrow 9}\equiv 0\ \pmod {3^2}$ since the exponent is certainly greater than $2$
  • $3^{4\uparrow 5\ldots\uparrow 9}\equiv 3^{(4\uparrow 5\ldots\uparrow 9)\ \bmod\ 28} \pmod{29}$
    • $28 = 2^2\times 7$
    • $4^{5\uparrow\ldots\uparrow 9}\equiv 0\pmod 4$
    • $4^{5\uparrow\ldots\uparrow 9}\equiv 4^{(5\uparrow\ldots\uparrow 9)\ \bmod\ 6}\pmod 7$
      • $5^{6\uparrow \ldots \uparrow 9} \equiv 1\pmod 6$ since the exponent is even
    • Thus, $4^{5\uparrow\ldots\uparrow 9}\equiv 4^1\equiv 4\pmod 7$
    • Chinese Remainder Theorem then says $4^{5\uparrow\ldots\uparrow 9}\equiv 4\pmod{28}$
  • Thus, $3^{4\uparrow 5\ldots\uparrow 9}\equiv 3^{4}\equiv 23\pmod {29}$
  • CRT says $3^{4\uparrow 5\ldots\uparrow 9}\equiv 81\pmod {522}$
• Thus, $2^{3\uparrow 4\uparrow 5\ldots\uparrow 9}\equiv 2^{81}\equiv 424\pmod {523}$

Second addend:

• $9^{8\uparrow 7\uparrow 6\ldots\uparrow 2}\equiv 9^{(8\uparrow 7\uparrow 6\ldots\uparrow 2)\ \bmod\ 522}\pmod {523}$
  • $522=2\times 3^2\times 29$
  • $8^{7\uparrow 6\ldots\uparrow 2}\equiv 0\pmod 2$
  • $8^{7\uparrow 6\ldots\uparrow 2}\equiv 2\pmod {3^2}$ since the exponent is even
  • $8^{7\uparrow 6\ldots\uparrow 2}\equiv 8^{(7\uparrow 6\ldots\uparrow 2)\ \bmod\ 28}\pmod {29}$
    • $28 = 2^2\times 7$
    • $7^{6\uparrow\ldots\uparrow 2}\equiv 1\pmod {2^2}$ since the exponent is even
    • $7^{6\uparrow\ldots\uparrow 2}\equiv 0\pmod 7$ (obvious)
    • CRT says $7^{6\uparrow\ldots\uparrow 2}\equiv 21\pmod {28}$
  • Thus, $8^{7\uparrow 6\ldots\uparrow 2}\equiv 8^{21}\equiv 12\pmod{29}$
  • CRT says $8^{7\uparrow 6\ldots\uparrow 2}\equiv 128\pmod{522}$
• Thus, $9^{8\uparrow 7\uparrow 6\ldots\uparrow 2}\equiv 9^{128}\equiv 99\pmod{523}$

Since $424+99=523$, the sum of the terms is multiple of $523$, just as we wanted to prove.