Modular arithmetic

[Stereotomy]

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?

[bo198214]

\( ^{n}a\text{ mod }b= {}^{m}a\text{ mod }b \)

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

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:
\( 7^7 \text{mod} 5 = 3\neq 2 = 7 \text{mod} 5 \)

[Stereotomy]

\( ^{n}a\text{ mod }b= {}^{m}a\text{ mod }b \)

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

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.

[bo198214]

\( 7^{7^{7}} \text{mod} 5 = 3 \)
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.