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I stumbled on this code, which finds the last digits of a tetration: https://blog.dreamshire.com/project-euler-188-solution/. The code seems to imply that $a^n \equiv a^{n \mod m} \mod m$ and takes advantage of this property to avoid calulation very large numbers. This implies that if you are working with powers (mod 10^m) you only need to care about the last m digits of the exponent.
I tried a few examples with pythons pow() and it seems to hold, but i cant find a proof. *with m > 1.

Wolfgang Pohrt
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    It seems to hold? Taking $m=10$, $n=12$, $a=2$ We have $12\equiv 2\pmod {10}$ so you are claiming that $2^{12}\equiv 2^2\pmod {10}$. Is that true? Or have I misread? – lulu Nov 23 '20 at 13:37
  • Note that this reduction does not help for tetrated expressions, since the exponent itself is huge. Far more important is that the power tower modulo some number becomes soon "stationary" meaning that from some number of entries on in the power tower the value does not change anymore. – Peter Nov 23 '20 at 13:39
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    The Python code is not "mod"ding the exponent. It is taking advantage of $a^n \bmod m \equiv (a \bmod m)^n$. – player3236 Nov 23 '20 at 13:41
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    Would you not expect that they need to change the modulus for each "height" in the tower? For example, for $a^b\pmod {10^8}$ they would need to find $b\pmod{\phi(10^8)}$ (provided $(a,10)=1$. Now as $b=a^c$ they would need to find $c\pmod{\phi(\phi(10^8))}$ etc. Shortly, it looks to me that it is wrong to take the same modulus at each stage. This is probably what bothers the OP too. –  Nov 23 '20 at 13:44
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    @Peter $3^{12}=531441,,3^2=9$. – lulu Nov 23 '20 at 13:45
  • OK, forget my comment – Peter Nov 23 '20 at 13:45
  • Im sorry i forgot to mention 10 by itself doesnt work. I dont think the code is taking advantage of $a^n \mod m \equiv (a \mod n)^n$. That identity couldnt be of any use since in the first place we want to calculate tha last digits of a tetration. Note that e.g. $2 \uparrow \uparrow 3$ is $2^{(2^2)}$ and not $(2^2)^2$. – Wolfgang Pohrt Nov 23 '20 at 15:33

2 Answers2

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This is valid for $m=10^8$. Namely, if $\gcd(a,10)=1$ we have (Euler's theorem):

  • $a^{2^7}\equiv 1\pmod{2^8}$ as $\varphi(2^8)=2^7$
  • $a^{4\cdot 5^7}\equiv 1\pmod{5^8}$ as $\varphi(5^8)=4\cdot 5^7$

Now, because both $2^7$ and $4\cdot 5^7$ are factors of $m=10^8$ we have that $a^m\equiv 1\pmod{2^8}$ and $a^m\equiv 1\pmod{5^8}$, so $a^m\equiv 1\pmod m$.

From this your statement follows fairly simply, but note that this is only valid for this very special modulus $10^8$, though the same proof would apply to all other powers of $10$, excluding $10$ itself. This is certainly not valid for a lot of other modules, as you have seen from the comments.

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Suppose $\,p = 2^{\large 2^{\Large k}}\!\!+1\,$ is a (Fermat) prime, e.g. $\,p=5\ (k=1)\,$ for the OP.

Then $\ a^{\large (2p)^{\large n}}\!\equiv 1\pmod{(2p)^n}\,$ if $\,\color{#0a0}{n\ge 2^k}\! = \log_2(p\!-\!1)\,$ and $\,(a,2p)=1\ $ since

$\!\bmod 2^n$ it's true by $\,\phi(2^n)\:=\: 2^{n-1}\mid\, (2p)^n,\,$ by Euler $\phi\,$ & $ $ MOR = modular order reduction.

$\!\bmod p^n$ it's true by $\,\phi(p^n)\! =\! 2^{2^{\large k}}\!p^{n-1}\mid (2p)^n\,$ by $\,\color{#0a0}{n\ge 2^k}\,$ with same justification as above.

Thus it's true mod their lcm $= (2p)^n$ by CCRT, so for $\,m\! =\! (2p)^n\,$ we have by MOR

$$\bbox[12px,border:1px solid #c00]{\bmod m\!:\,\ a^{m}\equiv 1\,\Rightarrow\, a^N\!\equiv a^{N\bmod m}}\qquad\qquad$$

Remark $ $ More generally see Carmichael's generalization of Euler's Theorem.

Bill Dubuque
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