# A base in which all primes end with $5$ different symbols?

In base $$10$$, all prime numbers (a part $$2$$ and $$5$$) end with $$1,3,7$$ or $$9$$, i.e. with four different symbols.

Is there a base in which all prime numbers end with $$5$$ different symbols (or also with $$5$$ distinct groups of symbols)? If yes, which base?

Thanks for your help! I apologize for such a trivial question!

NOTE: This question is related to this one.

• Is $5=\phi(n)$ for some $n$? – Lord Shark the Unknown Oct 19 '18 at 18:47
• Comments in the linked question advised you to study Euler's totient function $\phi(n).\,$ Did you do so before posting this question? If so, where are you stuck? If not, then you should do so. – Bill Dubuque Oct 19 '18 at 18:48
• Yes, but I did not understand much. I understood, as @LordSharktheUnknown said, that this is related to the totient function. But I don't know if I can evaluate all its values, since there is not a formula for the primes. Or what am I missing? – user559615 Oct 19 '18 at 18:52
• There is no $n$ such that $\phi(n)=5$, as $\phi(n)$ is even for $n>2$. – Inactive - Objecting Extremism Oct 19 '18 at 19:09
• @Servaes But sure! Many, many thanks! – user559615 Oct 19 '18 at 19:14

In base $$b$$, all prime numbers that do not divide $$b$$ end in a number between $$0$$ and $$b$$ coprime to $$b$$. Conversely, for every number between $$0$$ and $$b$$ coprime to $$b$$, there is a prime number ending in $$b$$. This follows from the prime number theorem for arithmetic progressions, for example. So the question becomes; for which numbers $$b$$ are there precisely $$5$$ numbers between $$0$$ and $$b$$ that are coprime to $$b$$. This number is denoted by $$\phi(b)$$, where $$b$$ is Euler's totient function. It is a simple result that $$\phi(b)$$ is even for all $$b>2$$, for example from the identity $$\phi\left(\prod_{i=1}^np_i^{k_i}\right)=\prod_{i=1}^np_i^{k_i-1}(p-1),$$ where the $$p_i$$ are distinct primes and the $$k_i$$ are positive integers.

In fact the only bases $$b$$ with $$\phi(b)<5$$ are $$b=2,3,4,5,6,8,10,12$$. Only in bases $$5$$ and $$8$$ do we get precisely $$5$$ different symbols in which primes can end, where we also count the divisors of the base.

• Thanks again! Also to the other users! – user559615 Oct 19 '18 at 19:16
• @AndreaPrunotto There is a simpler proof by reflection (negation) symmetry - see my answer. – Bill Dubuque Oct 19 '18 at 19:59

Hint  The number of residues coprime to $$n> 2$$ is even:  negation reflection $$\,x\mapsto -x\pmod {\!n}\,$$ partitions them into pairs (since it has no fixed points: $$\,-a\equiv a\,\Rightarrow\, n\mid 2a,\,$$ contra $$(n,a)=1)$$.

Remark $$\$$ Such use of reflections (or involutions) to pair-up terms frequently proves handy, e.g. see prior posts here on Wilson's theorem (in groups), esp. this one to start.

• Thanks for the clever answer! I have to study more!!!! – user559615 Oct 19 '18 at 20:52
• Just a little, further question. Is there a way to be sure that, for instance, $\phi(n)=10$ only for $n=10$ and $n=11$? – user559615 Oct 19 '18 at 22:24
• @AndreaPrunotto $\phi(10) = 4$ as you mention in your question. You can use the formula for $\phi$ for things like that (see many prior questions). – Bill Dubuque Oct 19 '18 at 22:51