How do I generalize the equation to be able to plug in any result for $\phi(n)=12$ and find any possible integer that works?

The following appeared in the problems section of the March 2015 issue of the American Mathematical Monthly. Show that there are infinitely many rational triples $(a, b, c)$ such that $a + b + c = abc = 6$. For example, here are two solutions $(1,2,3)$ and $(25/21,54/35,49/15)$. The deadline for submitting solutions was July […]

If $\operatorname{ord}_ma=10$, find $\operatorname{ord}_ma^6$. Justify you’re answer. I’m not sure what I should say for my answer to be justified. However, I expect $\operatorname{ord}_ma^6=5$, because $5\cdot 6=30$ and $10\mid 30$. Like I said, no idea if this is how to go about solving this problem.

This question already has an answer here: Fermat numbers and GCD 3 answers

How can we prove that $ p^{\frac1n} $ is irrational if $p $ is prime and $n>1$?

Is that true that all the prime numbers are of the form $6m \pm 1$ ? If so, can you please provide an example? Thanks in advance.

I am studying group theory so I do it by using the concept of group. What I am trying to prove is if p is prime then $(p-1)!\equiv-1\mod p$ Note that $\mathbb{Z_p}$ forms a multiplicative group. Hence $\forall a \in \{1,2,\dots,p-1\},\exists a^{-1}\in \{1,2,\dots,p-1\}$ This means that $aa^{-1}\equiv 1\mod p$ If $a=a^{-1}, 1 \equiv aa^{-1}= a^2 […]

I know the basics of logic, sets, relations and the like, so studying intros to abstract algebra and real analysis is not that hard. That said, I have a deficiency when it comes to elementary number theory and combinatorics topics. Things like divisibility, modular arithmetic, congruences, prime numbers and such like regularly pop up as […]

For my proof I distinguished the two possible cases which derive from $7 \mid a^2+b^2$: Case one: $7\mid a^2$ and $7 \mid b^2$ Case two (which (I think) is not possible): $7$ does not divide $a^2$ and $7$ does not divide $b^2$, but their sum. I’ve shown that case $1$ implies $7\mid a$ and $7\mid […]

how to prove $\forall p$ prime : $\binom{2p}{p} \equiv 2 \pmod p$ we have: $\binom{2p}{p} = \frac{2p (2p-1)(2p-3)…1}{p!p!}$ but how to continue?

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