Modular arithmetic - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: Modular arithmetic (/showthread.php?tid=435) Modular arithmetic - Stereotomy - 04/02/2010 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? RE: Modular arithmetic - bo198214 - 04/02/2010 (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: $7^7 \text{mod} 5 = 3\neq 2 = 7 \text{mod} 5$ RE: Modular arithmetic - Stereotomy - 04/03/2010 (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. RE: Modular arithmetic - bo198214 - 04/03/2010 (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.