# Tetration Forum

Full Version: Modular arithmetic
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I'll just preface this by saying I'm just a physics undergrad, so this might be a bit beyond my understanding, and I may well be missing something obvious or making a stupid mistake, but while playing around I noticed that it seems to be true that

$^{n}a\text{ mod }b= {}^{m}a\text{ mod }b$

$\text{For }^{n}a,{}^{m}a > b,\text{ }a,b,n,m \in \mathbb{N}$

Is this actually true? And if so is there a proof of it I'll be able to wrap my mind around?
(04/02/2010, 07:16 PM)Stereotomy Wrote: [ -> ]$^{n}a\text{ mod }b= {}^{m}a\text{ mod }b$

$\text{For }^{n}a,{}^{m}a > b,\text{ }a,b,n,m \in \mathbb{N}$

Is this actually true? And if so is there a proof of it I'll be able to wrap my mind around?

I dont think it is true. For example:
$7^7 \text{mod} 5 = 3\neq 2 = 7 \text{mod} 5$
(04/02/2010, 10:57 PM)bo198214 Wrote: [ -> ]
(04/02/2010, 07:16 PM)Stereotomy Wrote: [ -> ]$^{n}a\text{ mod }b= {}^{m}a\text{ mod }b$

$\text{For }^{n}a,{}^{m}a > b,\text{ }a,b,n,m \in \mathbb{N}$

Is this actually true? And if so is there a proof of it I'll be able to wrap my mind around?

I dont think it is true. For example:
$7^7 \text{mod} 5 = 3\neq 2 = 7 \text{mod} 5$

Ah, good point, though
$7^{7^{7}} \text{mod} 5 = 3$
Is true as well. In fact, thinking about it, the numbers I tried out with this all had b>a. Perhaps that's an additional condition that either b > a or m, n > 1?

Just quickly tried this for a few low examples, a = 8, 9, 10, 11, and it seems to hold.
(04/03/2010, 12:18 AM)Stereotomy Wrote: [ -> ]$7^{7^{7}} \text{mod} 5 = 3$
Is true as well. In fact, thinking about it, the numbers I tried out with this all had b>a. Perhaps that's an additional condition that either b > a or m, n > 1?

There is an article which proves that ${^n a}$ (which is $a{\uparrow}^2 n$ in Knuth's arrow notation) finally will be constant for $n\to\infty$ mod any $M$, see this thread.