# Are there infinitely many $c$-th perfect powers having a constant congruence speed of $c$?

Let $$a$$ be a positive integer of the form $$20 \cdot n + 5$$ (i.e., $$a : a \equiv 5 \pmod {20}$$, $$n \in \mathbb{N}_0$$). I wish to prove (or disporove) the following statement.

Let $$c \in \mathbb{Z}^+$$ be given. Then, $$\forall c$$, $$\exists^{\infty} n:=n(c)$$ such that $$a$$ is a perfect $$c$$-th power of an integer (i.e., $$\sqrt[c]{20 \cdot n + 5} \in \mathbb{N}$$) and such that $$\left(2^c \mid (20 \cdot n + 4) \wedge 2^{c+1} \nmid (20 \cdot n + 4) \right)$$.

This result would be a sufficient but not necessary condition to prove the existence of infinitely many $$c$$-th perfect powers characterized by a constant congruence speed of $$c$$ [see Equation (16) of "Number of stable digits of any integer tetration”, NNTDM, 28(3), p. 454], since here we are just taking into account the fifth line of the aforementioned (16), but we could use a very similar argument also for lines $$3$$, $$4$$, $$6$$, and so forth.

As an example, let $$\tilde{c}:=4$$ so that we are only taking into account the perfect squares of squares.
Then, we have that $$a=20 \cdot n + 5 \Rightarrow a^4 - 1 = 160000 \cdot n^4 + 160000 \cdot n^3 + 60000 \cdot n^2 + 10000 \cdot n + 624$$.
Now, $$\frac{2^4(10000 \cdot n^4 + 10000 \cdot n^3 + 3750 \cdot n^2 + 625 \cdot n + 39)}{2^\tilde{c}} \in \mathbb{Z}^+$$ clearly holds for infinitely many $$n$$, since we only need to take $$\tilde{a}:=10000 \cdot n^4 + 10000 \cdot n^3 + 3750 \cdot n^2 + 625 \cdot n + 39$$. On the other hand, we have that $$\frac{2^4(10000 \cdot n^4 + 10000 \cdot n^3 + 3750 \cdot n^2 + 625 \cdot n + 39)}{2^{\tilde{c}+1}} \Rightarrow \frac{10000 \cdot n^4 + 10000 \cdot n^3 + 3750 \cdot n^2 + 625 \cdot n + 39}{2}$$ is not integer if and only if $$n$$ is even (since $$10000 \cdot n^4 + 10000 \cdot n^3 + 3750 \cdot n^2$$ is always even and $$625 \cdot n + 39$$ is odd if and only if $$625 \cdot n$$ is even).

By Equation (16), if $$n$$ is even, the constant congruence speed of these tetration bases $$\tilde{a}$$ is exactly $$4$$ (i.e., $$V(\tilde{a})=\tilde{c}$$ by construction).

I don't have an answer to the problem, but I've found the smallest $$n$$ values that satisfy the conditions for $$3≤c≤32$$:

c = 3, n = 781
c = 4, n = 31
c = 5, n = 28832519531
c = 6, n = 36754594531
c = 7, n = 62689544011144531
c = 8, n = 331710215644531
c = 9, n = 4890715999471794558542460644531
c = 10, n = 3631230373943953665696410644531
c = 11, n = 23589066992907634595308472834860910644531
c = 12, n = 67244441212314921858906745910644531
c = 13, n = 3625870143853203819926363508657598642713511005476495910644531
c = 14, n = 1465140843884714270976148969202976910924406722187995910644531
c = 15, n = 38690479963498809091737863348793164420935551529906904548065662995910644531
c = 16, n = 312631332647556581305305681592478767196748312995910644531
c = 17, n = 11569470373550244296710575636413735557980570633179868742716598621095439617347781541027812995910644531
c = 18, n = 2525290715570784335046346083455593617243861635810431846648502966903199377587758100662812995910644531
c = 19, n = 272965352122948423752826950130428069320838745772671674235018873992912165628578321676375280800790444112812995910644531
c = 20, n = 117436088442337833299014285033231187408850519625447518040999705715950565729411547587677374612812995910644531
c = 21, n = 158583337523427141564545626841668185936539034747454077706101819086145762621004853283309408410418948993676596769437535035798411784819612812995910644531
c = 22, n = 18685042339714006477754576530273613178408543951793163484775085983988697940571819122547886999378525176823360877417409126044983018369612812995910644531
c = 23, n = 8272313144106030414239803015112164873941113539123827520294752496641674175800170286994891390941445559934133075393915224247119850429249292297129159123869612812995910644531
c = 24, n = 739598932335862574122742576863133113342625684931126885858020880431445980172485199105489438212900307394663451669869838433249082467127308314548869612812995910644531
c = 25, n = 9336161182574906263873577625864167609976250329802240672844277874632395850026033164714145215137284594914036005724957972191300983669445499538367929712573980231210444004312170034332041268498869612812995910644531
c = 26, n = 593771975901882112678470120072180834091768391371150159228148474527843005983870725060549974836390387720104768639425511425466357979396679573763405633621462317284586052271263551504805175998869612812995910644531
c = 27, n = 1076739759255702964443403608095182515794218024272299236506423095481325867740849093672759550307068030053531051268489586131274344836666343118294763759327886481411819732001406179029535901190292275831892577338380998869612812995910644531
c = 28, n = 70600378424472411784637196437064020201285204406753217321176406856980969058621673032867984964865217975074384951947012176282449075075206272174426972294066228104034061221963316124451480872970280830998869612812995910644531
c = 29, n = 2360693050339016440612713148634556653265977115036079573465918767988631170065902810025770690123896495680297431111387181673528312933504544286079211370886692749683010577826688336929058499911205503965708164855649737865166432095946001360877058569266867960330998869612812995910644531
c = 30, n = 81040827236259886486647152297190590600290791352489256598028395718738135098340153550012698994819091894940130045531095865939758268823947116131985237854880688616823962574669344604731697718940294386418118407649416597488180662133096284119329632397502755330998869612812995910644531
c = 31, n = 601942123223942429715861981956164423460485793673539969191506846415037418354429114742623024308014213287422922071017344877608372739249661394370337673791835717428567171170385814123596540871455283924699152201294419018461297714848865197887998213950022192225271009840595489253746705330998869612812995910644531
c = 32, n = 11396367448074787398620886723024893477194385311819608773878657515474676336937054201088251900401807243384712603574012901876429995732597527372933683845882700295794545240354609458836695990318038781303848255015176323038369983069777358454331257324505205330998869612812995910644531


You can see that the last digits of each number are converging to $$...05330998869612812995910644531$$.

This is the Mathematica code that produced these results:

cMax = 32;
resultList = {};
For[c = 3, c <= cMax, c++, k = 5;
foundSolution = False;
While[! foundSolution, potentialN = (k^c - 5)/20;
If[IntegerQ[potentialN] && Mod[20*potentialN + 4, 2^c] == 0 &&
Mod[20*potentialN + 4, 2^(c + 1)] != 0,
AppendTo[resultList, {c, potentialN}];
foundSolution = True; Print[c, ": ", potentialN]];
k += 5;]]
resultList

• Thank you for sharing this. Here, we simply need to multiply those numbers by $20$ and then just add $5$. So, we'll get $c=3 \Rightarrow 15625$, $c=4 \Rightarrow 625$, $c=5 \Rightarrow576650390625$, $c=6 \Rightarrow735091890625$, $c=7 \Rightarrow1253790880222890625$, $c=8 \Rightarrow6634204312890625$, $c=9 \Rightarrow97814319989435891170849212890625$, $\dots$, converging to the $10$-adic integer $\alpha_{25}$ by Def. 2.6 of the mentioned paper, which is one of the $15$ solutions of the fundamental equation $y^5=y$ in the commutative ring of $10$-adic integers. May 30 at 17:55
• Now, since $\alpha_{25}=\{5^{2^n}\}_{\infty}=\dots 92256259918212890625$ is not limited on the left, we have just shown the existence of solutions for any given value of $c \in \mathbb{N}-\{0,1\}$, while the $c=1$ case is trivial by taking another base from another $10$-adic solution of the same equation, such as $a=2$, and that's all. Will somebody contribute with a final proof of the stated conjecture (since we are almost there)? May 30 at 18:22

Ok, I have just found a way to prove (a generalization) of the present statement.

I am planning to write a preprint in a few of weeks, then I will be happy to share it here.

Basically, it is just a consequence of https://arxiv.org/pdf/2208.02622.pdf, Equation 2, line 6. So, it is sufficient to play a couple of little math tricks, such as adding/subtracting $$1$$ to the $$(c+1)$$-th rightmost digit of the string $$\dots 18212890625$$ to be sure that $$10^n$$ multiplied by that string of $$c+1$$ digits will be characterized by the same constant congruence speed (equal to $$c$$) for any $$n \in \mathbb{N}_0$$, and then we only need to invoke the well-known Hensel's (lifting) lemma in order complete a constructive proof of this little corollary of a more general theorem, rising every element of the described set to the $$c$$-th power as requested (here we are using the property that $$a \equiv \dots 18212890625 \pmod {10^c} \Rightarrow a^n \equiv \dots 18212890625 \pmod {10^c}$$ holds for any positive integer $$n$$). Q.E.D.

• The proof above is trivially complete after we note that $a=2$ is not a perfect power (i.e., a perfect square, a perfect cube, and so forth) and has a unitary constant congruence speed, while the constant congruence speed of $32$ is $2$. Thus, let $x=42$ and it follows that $10^n \cdot 42$ is a solution for any $n \in \mathbb{N}$. May 30 at 23:37
• At the end, I have proven the result I was looking for by simply taking into account line 7 of the mentioned Equation (16) (arxiv.org/abs/2210.07956), but the same outcome is also true by considering lines 4&5. As promised, here is a preprint of mine that strictly proves the whole conjecture stated in this thread: hal.science/hal-04131657v1 Jul 3 at 23:48