Solved with franchester, ingenio, tworigami, GameMaster402, anish9876, peeyushmaths, AIME12345, SD2014, AOPS12142015, huangyi_99, sriraamster, kothasuhas, budu, and mathfun5. Proceed w/ principle of presuming paradox, and producing preposterous products. Without perdition of principle, presume the smallest power of phive is the primary power of the progression ($a_{1}^{2018}+a_{2}$). Propose a positive $p$ performing $5^{p}={a_1}^{2018}+a_2$. By the presentation of $p$, we perceive that $5^p$ proportionately partitions all parts of the progression. The pivotal proposition proceeds: Proposition: $5^p$ proportionately parts $a_1^{2018^{2018}}+a_1$. Proof: Perform modulo $p^{\text{th}}$ power of phive. Our party personally presents: $${a_1}^{2018^{2018}} \equiv {a_2}^{2018^{2017}} \equiv {a_3}^{2018^{2016}} \ldots \equiv {a_{2018}}^{2018} \equiv -a_1 \pmod{5^p}$$as preferred. Practicing Pierre de Fermat's petite property, the precipitate in modulo $5$, $${a_1}^{2018^{2018}}+a_{1} \equiv {a_1}^{4}+a_1 \equiv a_1({a_1}^3+1) \pmod{5},$$purporting $a_1 \equiv 0,-1 \pmod{5}$. Provoking LTE, since $2018^{2018}-1 \equiv 3 \pmod{5}$, $$p \leq v_p({a_1}^{2018^{2018}}+a_{1}) \leq v_p(a_1) + v_p(a_1+1) \Rightarrow {a_1}^{2018}+a_2 \leq 5^p\leq a_1+1.$$The partisanship proves $a_{1}=a_2=1$, proving $2=5^p$, a preposterous predicament, proving the proposition. $\blacksquare$

mira74 wrote: Pierre de Fermat's petite property im literally dying from fermat's little theorem

At first of all we must look into case when $5\mid a_i$ for some $1\le i\le 2018$. My claim is that $5$ do not devide any $a_i$ . Suppose $5$ devides some of $a_i$ .Also assume $v_5(a_k)=\max \{v_5(a_1),\cdots ,v_5(a_{2018})\}$. Hence $a_k^{2018} +a_{k+1}$ Is not a perfect power of $5$ which is a contradiction !! Now note that $a_1^{2018} \equiv -a_2 \pmod{5}$. \[\Rightarrow a_1^{2018^2} +a_3 \equiv a_{2}^{2018}+a_3 \pmod{5}\] \[ \Rightarrow a_1^{2018^{2018}} +a_1 = a_{2018}^{2018}+a_1\equiv 0 \pmod{5}\] Since $5$ do not devide $a_1$ hence \[ 5\mid a_1^{2018^{2018}-1} +1\]which forces \[5\mid a_1^{2.2018^{2018} -2} -1\] Since $a_1^4\equiv 1\pmod{5}$ it means $4\mid 2.2018^{2018}-2$ . Clearly It is a contradiction.

Suppose not. In what follows, all indices are taken modulo $2018$. Note that if $\nu_5(a_i^{2018}) \neq \nu_5(a_{i + 1})$ for any $i$, we have an immediate contradiction, as then $a_i^{2018} + a_{i + 1}$ will be divisible by a prime other than $5$. Hence, $2018\nu_5(a_i) = \nu_5(a_{i + 1})$ for all $i$. Iterating this yields \[ \nu_5(a_1) = 2018\nu_5(a_{2018}) = 2018^2\nu_5(a_{2017}) = \cdots = 2018^{2018}\nu_5(a_1), \]so $\nu_5(a_1) = 0$, implying that $\nu_5(a_i) = 0$ for all $i$. Now, note that we must have $a_{i + 1} \equiv -a_i^{2018} \pmod{5}$. Iterating this yields \begin{align*} a_1 \equiv -a_{2018}^{2018} \equiv -a_{2017}^{2018^2} \equiv \cdots \equiv -a_1^{2018^{2018}} \pmod {5}. \end{align*}Since $a_1 \not\equiv 0 \pmod{5}$, we have $a_1^{2018^{2018}} \equiv 1 \pmod{5}$, so $a_1 \equiv -1 \pmod{5}$. This is enough to imply that $a_i \equiv -1 \pmod{5}$ for all $i$. Now, WLOG suppose $a_1^{2018} + a_2$ is minimal. Then $a_1^{2018} + a_2$ divides all of the other terms, so \begin{align*} a_{i + 1} &\equiv -a_i^{2018} \pmod{a_1^{2018} + a_2}. \end{align*}Iterating this yields \begin{align*} a_1 &\equiv -a_1^{2018^{2018}} \pmod{a_1^{2018} + a_2} \\ a_1^{2018^{2018} - 1} + 1 &\equiv 0 \pmod{a_1^{2018} + a_2}, \end{align*}where dividing by $a_1$ makes sense because $\gcd(a_1, a_1^{2018} + a_2) = \gcd(a_1, a_2) = 1$ (if $\gcd(a_1, a_2) > 1$, then $a_1^{2018} + a_2$ would be divisible by a prime other than $5$, as $5 \nmid a_1, a_2$). Recalling that $5 \mid a_1 + 1$, we have from LTE that \begin{align*} \nu_5\left(a_1^{2018^{2018} - 1} + 1\right) &= \nu_5(a_1 + 1) + \nu_5(2018^{2018} - 1). \end{align*}However, $2018^{2018} \equiv 3^{2018} \equiv 3^2 \equiv 9 \pmod{5}$, so $\nu_5\left(a_1^{2018^{2018} - 1} + 1\right) = \nu_5(a_1 + 1)$. Therefore, since $a_1^{2018} + a_2$ is a power of $5$, we have \begin{align*} a_1^{2018} + a_2 &\mid a_1 + 1. \end{align*}Hence, we must have $a_1^{2018} + a_2 \leq a_1 + 1$, but $a_1^{2018} > a_1$ and $a_2 > 1$ (since $a_1 \equiv a_2 \equiv 4 \pmod{5}$). This is a contradiction, so we are done. $\Box$

Solution from Twitch Solves ISL: Assume the contrary. Claim: None of the numbers are divisible by $5$. Proof. If $\nu_5(a_1)$ is maximal and positive, then $\nu_5(a_1^{2018}) > \nu_5(a_2)$, so $a_1^{2018} + a_2$ cannot be a power of $5$. $\blacksquare$ Let $5^e$ denote the smallest of the powers of $5$, so all elements are divisible by $5^e$. Evidently, $e \ge 2$, and for every index $i$, \[ a_{i+2} \equiv - a_i^{2018^2} \pmod{5^e} \] Claim: We have $a_i \equiv 4 \pmod 5$ for all $i$. Proof. This follows immediately form the previous displayed equation. $\blacksquare$ Now let $x = -a_1 \pmod{5^e}$. But if we iterate this $1009$ times we obtain \[ x \equiv x^{2018^{2018}} \pmod{5^e} \]Thus $x \bmod{5^e}$ has order dividing $N = 2018^{2018}-1$. But since $N \equiv 3 \pmod 5$, we find $\gcd(N, \varphi(5^e)) = 1$, So this forces $x \equiv 1$. In other words, $a_1 \equiv -1 \pmod{5^e}$, and the same is true for any other term. But $(5^e-1)^{2018} > 5^e$ and we have a size contradiction.

Cool problem. We proceed by contradiction. Claim: $\nu_5(a_i) = 0$ for all $i$. Proof: Suppose the contrary, and let $\nu_5(a_1) \geq 1$ be maximal. It is evident that $\nu_5(a_1^{2018}) > \nu_5(a_2)$, a contradiction. Remark that $a_i^{2018^{2018}} \equiv -a_i \pmod 5,$ and as $a_i$ is not divisible by $5$, we must have $a_i \equiv -1 \pmod 5$ for all $i$. Now suppose $\nu_5(a_1^{2018}+a_2) = k$ is minimal, so $$a_1^{2018^{2018}-1} + 1 \equiv 0 \pmod {5^k}.$$Now we are almost done; note that $$\nu_5(a_1^{2018^{2018}-1} + 1) = \nu_5(2018^{2018}-1) + \nu_5(a_1+1) = \nu_5(a_1+1),$$so $a_1^{2018}+a_2 \leq a_1+1$ which is obviously false.

Kayak wrote: Show that there do not exist natural numbes $a_1, a_2, \cdots, a_{2018}$ such that the numbers \[ (a_1)^{2018}+a_2, (a_2)^{2018}+a_3, \cdots, (a_{2018})^{2018}+a_1 \]are all powers of $5$ Proposed by Tejaswi Navilarekallu Assume to the contrary there exists such numbers. Claim : $5$ does not divide $a_i$ for any $a_i$ Proof : Clearly if $5$ divides one of $a_i$ it divides all.Now consider the number $a_k^{2018}+a_{k+1}$ where $k$ is chosen so that $\nu_5(a_k)=\max{\nu_5(a_i)}$. Clearly $\nu_5(a_k^{2018}+a_{k+1})$ cant be a power of $5$ since $\nu_5(a_k^{2018}) > \nu_5(a_{k+1}) >0 $ which concludes the claim.$\square$ Next note that \begin{align*} a_1^{2018}\equiv -a_2\pmod 5 \\ a_2^{2018}\equiv -a_3\pmod 5 \\ \cdots a_{2018}^{2018}\equiv -a_1\pmod 5 \end{align*}Substituting repeatedly we get $a_1^{2018^{2018}}\equiv -a_1\pmod 5$.Thus $a_1^{2(2018^{2018}-1)}\equiv 1 \pmod 5$.Thus $5-1\mid 2(2018^{2018}-1)$.But this is a contradiction.$\blacksquare$

Claim: We have $a_i\not \equiv 0 \pmod5$ for all $i$. Proof: Note the following fact: for positive integers $x$ and $y$, if $x+y$ is a power of $5$, then $\nu_5(x)=\nu_5(y)$. Therefore, since $a_i^{2018}+a_{i+1}$ is a power of $5$ for each $i$, we have $\nu_5(a_{i+1}) = \nu_5(a_i^{2018}) = 2018\nu_5(a_i)$. Iterating this, we have $\nu_5(a_1)=2018^{2018}\nu_5(a_1)$ since the sequence cycles back, which means $\nu_5(a_1)=0$, and hence $\nu_5(a_i)=0$ for all $i$. Therefore, $5\nmid a_i$ for all $i$, as claimed. $\blacksquare$ Let $5^m$ be the smallest power of $5$ in the list, WLOG it is $a_1^{2018}+a_2$. (We can WLOG since the variables are cyclically symmetric.) Then all of the elements of the list are multiples of $5^m$. We have for all $i$ (indices cycle) that $a_i^{2018}+a_{i+1} \equiv 0 \pmod{5^m}$, so \[-a_{i+1}\equiv (-a_i)^{2018} \pmod{5^m}.\]Iterating this, we have $(-a_1)^{2018^{2018}} \equiv -a_1 \pmod{5^m}$. Since $5\nmid a_1$, we have \[(-a_1)^{2018^{2018}-1} \equiv 1 \pmod{5^m}. \qquad (\clubsuit) \]In particular, said equivalence also holds in $\pmod5$. Since $2018^{2018}-1\equiv 3\pmod4$, this implies $(-a_1)^3\equiv 1\pmod5$, so $a_1\equiv -1\pmod5$. Now, $(\clubsuit)$ implies \[ m\le \nu_5\left( a_1^{2018^{2018}-1}+1\right) = \nu_5(a_1+1) + \nu_5(2018^{2018}-1) = \nu_5(a_1+1)\]by LTE. Therefore, $5^m\mid a_1+1$, but $5^m =a_1^{2018}+a_2$, a size contradiction.

Isn't This Easy for INDIA TST ??

MatBoy-123 wrote: Isn't This Easy for INDIA TST ?? Nope, it's not.

Kayak wrote: Show that there do not exist natural numbes $a_1, a_2, \cdots, a_{2018}$ such that the numbers \[ (a_1)^{2018}+a_2, (a_2)^{2018}+a_3, \cdots, (a_{2018})^{2018}+a_1 \]are all powers of $5$ Proposed by Tejaswi Navilarekallu Nice Problem. FTSOC Assume the existence of such sequence of natural numbers. Now just observe if for any $i\leq 2018$ we have $a_i\equiv 0\mod 5$ then all the terms of sequence $(a_k)_{k=1}^{2018}$ is divisible by $5$ Now let's assume such case when all the terms are divisible by $5$. Then we will Prove that this is never possible. For it let's start with $5|a_1$ then observe as $a_1^{2018}+a_2=5^{b_1}\implies 5^{2018}|a_2$ Similarly we can get $5^{2018^{2018}}|a_3,...$ and so on. But then $a_{2018}^{2018}+a_1=5^{b_{2018}}$ will be the clear contradiction. Now just observe if $a_1^{2018}+a_2=5^{b_1}$ is the smallest possible power of $5$ in this then we will have $a_1^{2018}\equiv -a_2\mod 5^{b_1},a_2^{2018}\equiv -a_3\mod 5^{b_1},...a_{2018}^{2018}\equiv -a_1\mod 5^{b_1}\implies a_1^{2018^{2018}}\equiv -a_1\mod 5^{b_1}$ So $a_1^{2(2018^{2018}-1)}\equiv 1\mod 5^{b_1}$ and as $gcd(2(2018^{2018}-1),\varphi(5^{b_1}))=1$ So we must have $a_1\equiv -1\mod 5^{b_1}$ now this is enough to prove the same for all $a_i's$ and Hence we will get Size Contradiction by LTE. So we are done there doesn't exist such sequence of natural numbers. $\blacksquare$

niyu wrote: \begin{align*} a_1 \equiv -a_{2018}^{2018} \equiv -a_{2017}^{2018^2} \equiv \cdots \equiv -a_1^{2018^{2018}} \pmod {5}. \end{align*}Since $a_1 \not\equiv 0 \pmod{5}$, we have $a_1^{2018^{2018}} \equiv 1 \pmod{5}$, so $a_1 \equiv -1 \pmod{5}$. This is enough to imply that $a_i \equiv -1 \pmod{5}$ for all $i$. Why $a_1^{2018^{2018}}+a_1\equiv 0 \pmod{5}$ means $a_1^{2018^{2018}} \equiv 1 \pmod{5}$?

CROWmatician wrote: niyu wrote: \begin{align*} a_1 \equiv -a_{2018}^{2018} \equiv -a_{2017}^{2018^2} \equiv \cdots \equiv -a_1^{2018^{2018}} \pmod {5}. \end{align*}Since $a_1 \not\equiv 0 \pmod{5}$, we have $a_1^{2018^{2018}} \equiv 1 \pmod{5}$, so $a_1 \equiv -1 \pmod{5}$. This is enough to imply that $a_i \equiv -1 \pmod{5}$ for all $i$. Why $a_1^{2018^{2018}}+a_1\equiv 0 \pmod{5}$ means $a_1^{2018^{2018}} \equiv 1 \pmod{5}$? bbuummpp

New finish? Solution. With$\pmod{5}$ we get $a_1 \equiv \cdots \equiv a_{2018} \equiv \{0,-1\} \pmod{5}$. So firstly, assume $a_1 \equiv 0 \pmod{5}$. Let $i_0$ be the index such that $\nu_5(a_{i_0}) = \displaystyle\max_{i \in \mathbb{Z}}\nu_5(a_i)$. Now notice that, \[ a_{i_0}^{2018} + a_{i_0+1} = 5^{k_{i_0}}, \]we know $2018\cdot\nu_5(a_{i_0}) > \nu_5(a_{i_0 + 1})$, thus $k_{i_0} = \nu_5(a_{i_0+1})$, but because of size reasons this is a clear contradiction. Otherwise, $a_1 \equiv \cdots \equiv a_{2018} \equiv -1 \pmod{5}$. Perform a change of variables $5b_i-1=a_i$. \begin{align*} &\implies(5b_j-1)^{2018}+(5b_{j+1}-1) \\ &= (5b_{j}-1)^{2018}-1 + 5b_{j+1} \\ &= \left((5b_{j})^{2018} + \left(\sum_{i=1}^{2016} \binom{2018}{i}(-1)^{i}(5b_{j})^{2018-i} \right) - 2018\cdot5b_j \right) + 5b_{j+1} \end{align*} If there exists any $j$ such that $\nu_5(5b_j) > \nu_5(5b_{j+1})$ then we are done because of size reasons. So, $\nu_5(5b_1) \leq \nu_5(5b_2) \leq \cdots \leq \nu_5(5b_{2018}) \leq \nu_5(5b_{1})$, which means all of them are equal, making another change of variables $5b_i = 5^\alpha \cdot c_i, \gcd(5,c_i)=1$ We get, \begin{align*} &\implies(5b_i-1)^{2018} + (5b_{i+1}-1)\\ &= \left((5^{\alpha}c_i)^{2018} + \left(\sum_{j=1}^{2016} \binom{2018}{j}(-1)^{j}(5^{\alpha}c_{i})^{2018-j}\right) - 2018\cdot 5^{\alpha}c_j \right) + 5^{\alpha}c_{j+1} \\ &\equiv -2018 \cdot 5^{\alpha}c_j + 5^{\alpha}c_{j+1} \pmod{5^{\alpha+1}} \\ &\implies -2018 \cdot c_j + c_{j+1} \equiv 2 c_j + c_{j+1} \equiv 0 \pmod{5} \end{align*}as otherwise we are done because of size reasons. Thus, \[c_2 \equiv -2c_1, c_3 \equiv 4c_1, \cdots, c_{2018} \equiv (-2)^{2017}c_1 \implies c_1 \equiv 2^{2018}c_1 \pmod{5}.\]From which we get $5|c_1$, a contradiction.

Funny problem! Assume the contrary, i.e that all of them are powers of $5$. Now if one of the $a_i$ is divisible by $5$, then so are the rest, and if WLOG $max(v_5(a_i))=v_5(a_1)$ we have a contradiction with $(a_1)^{2018}+a_2$ being a power of $5$ $\Rightarrow $ none of the $a$'s is divisible by $5$. Now assume that WLOG $(a_1)^{2018}+a_2=5^x$ is the smallest out of all of those powers of $5$,so it divides all of the other numbers. We get that $$ 5^x|(a_2)^{2018}+a_3=(5^x-(a_1)^{2018})^{2018}+a_3 \Leftrightarrow 5^x|(a_1)^{2018^2}+a_3$$thus $a_3\equiv - (a_1)^{2018^2}(mod 5^x)$. Continuing in this fashion by induction we can get that $a_{n}\equiv - (a_1)^{2018^{n-1}}(mod 5^x)$ and thus $$(a_{2018})^{2018}+a_1\equiv (a_1)^{2018^{2018}}+a_1=a_1((a_1)^{2018^{2018}-1}+1)\equiv 0(mod 5^x)$$so $a_1^{2018^{2018}-1}\equiv -1(mod 5^x)$. If $ord_{5^x}(a_1)=d$, then $d|gcd(2(2018^{2018}-1);4.5^{x-1})=2$, so either way $a_1^2\equiv 1(mod 5^x)$. This however is impossible if $a_1>1$ because $5^x=(a_1)^{2018}+a_2>(a_1)^2>5^x$, so we must have $a_1=1$. But then $(a_{2018})^2018 + a_1\equiv (a_1)^{2018^{2018}}+a_1 = 2 \equiv 0(mod 5^x)$, which isn't possible, so the system has indeed no solutions!

Solved with Rama 1728 First assume by contradiction that there exists such $a_1,a_2,...,a_{2018}$ Claim 1: $a_k+1 \equiv 0 \pmod 5$ Proof: First we let $a_{2018+b}=a_{b}$ and now note that by FLT we have $a_{k-1}^2+a_k \equiv 0 \pmod 5$ hence since $a_{k-1}^2$ is $-1,0,1$ on mod 5 we get that all $a_k$ are $-1,0,1$ on mod 5. Case 1: There exists at least one $a_n$ multiple of $5$. W.L.O.G let $a_1$ be the one with $v_5(a_1)$ maximun compared with the others hence let $a_1^{2018}+a_2=5^k$ and now using $v_5$ we have that $v_5(a_2)=m$ which is not possible since $v_5(a_1)<m$. Case 2: There exists $a_n$ such that $a_n \equiv 1 \pmod 5$ Then we get that $a_{n+1} \equiv -1 \pmod 5$ and note that its a recursion and in some moment we will get $a_n \equiv -1 \pmod 5$ which is not possible Hence we get that all $a_k$ are $-1$ in mod 5. Claim 2: $a_k^{2018^{2018}-1}+1 \equiv 0 \pmod 5^{c}$ where $5^c$ is the smallest power of $5$ between $a_1^{2018}+a_2,...,a_{2018}^{2018}+a_1$ Proof: First W.L.O.G let $a_1^{2018}+a_2=5^c$. It suffices to show that $a_k^{2018^{2018}}+a_k \equiv 0 \pmod 5^c$ and now note that by the conditions. $$a_k^{2018^{2018}} \equiv a_{k+1}^{2018^{2017}} \equiv ... \equiv a_{k-1}^{2} \equiv -a_k \pmod 5^c$$Hence proved!. Final proof: By LTE using Claim 1 and Claim 2 we have that. $$c \le v_5(a_k^{2018^{2018}-1}+1)=v_5(a_k+1)+v_5(2018^{2018}-1)=v_5(a_k+1) \implies a_1^{2018}+a_2 \le a_k+1$$Setting $k=1$ we get that $a_1=a_2=1$ but that means $2$ is a power of $5$ which is a contradiction!! Hence no such $a_1,a_2,...,a_{2018}$ exists! (thus we are done )

Assume contrary. Note none of the $a_i$ should be divisible by $5$, otherwise we $a_k^{2018} + a_{k+1}$ is not a power of $5$ where $v_5(a_k) =\max \{ v_5(a_1),v_5(a_2),\ldots,v_5(a_{2018}) \}$. Then using FLT we get $$a_3 \equiv -a_2^{2018} \equiv - \left(-a_1^{2018} \right)^{2018} = - a_1^{\text{a multiple of }4} \equiv -1 \pmod{5} $$So each $a_i$ is $-1$ modulo $5$. Write $a_i = 5b_i - 1$. Let $v_5( \gcd(b_1,\ldots,b_{2018})) = e-1$, with $e \ge 1$. Write each $b_i = 5^{e-1} \cdot c_i$. Then $a_i = 5^e \cdot c_i - 1$. By definition not all $c_i$ are divisible by $5$. Now note each power of $5$ is at least $5^{e+1}$ (as $(5^e - 1)^{2018} > 5^e$). Hence, \begin{align*} 0 &\equiv a_i^{2018} + a_{i+1} = (5^e \cdot c_i - 1)^{2018} + 5^e c_{i+1} - 1 \equiv \binom{2018}{1} \cdot (-1)(5^e c_i) + 1 + 5^e c_{i+1} -1\\ &\equiv -3 (5^e \cdot c_i) + 5^e c_{i+1} = 5^e( c_{i+1} - 3c_i) \pmod{5^{e+1}} \\ &\qquad \implies c_{i+1} \equiv 3c_i \pmod{5} \qquad \qquad (1) \end{align*}Using $(1)$ repeatedly gives $$ c_i \equiv 3c_{i-1} \equiv 9c_{i-2} \equiv \cdots \equiv 3^{2018} \cdot c_i \equiv 3^2 \cdot c_i \equiv -c_i \pmod{5} $$But this means $5$ divides every $c_i$, contradiction. $\blacksquare$

Funny problem! FTSOC. First see the sequences in module $5$, we have two cases: Case 1: If $a_i \equiv 0\pmod 5$ This case is east only choose the maximum of $v_{5}(a_i)$ so $2018v_{5}(a_{i})>v_{5}(a_{i+1})$ then $v_5(a_{i}^{2018} + a_{i+1})=v_5(a_{i+1})$ this obvious is contradiction because $a_{i}^{2018} + a_{i+1}>a_{i+1}$ so $a_{i}^{2018} + a_{i+1}$ isn't a power of $5$ Case 2: If $a_i \equiv -1\pmod 5$ Let $a_i +1=5^{k_i}.b_i$ with $k_i=v_5(a_i)$. We rewrite $a_{i}^{2018}-1 + a_{i+1}+1$ is a power of 5 then by LTE $v_5(a_{i}^{2018}-1 )=k_i$ so $k_i=k_{i+1}$ (otherwise the contradiction is the same like the case 1) but: $$\frac{a_{i}^{2018}-1}{a_i +1} = (a_i -1)( a_i^{2016}+a_i^{2014}+\dots +1) \equiv 2 \pmod 5$$If we divide all original expressions by $5^{k_i}$ we have that the new expressions are multiples of $5$, but: $$\frac{a_{i}^{2018}-1 + a_{i+1}+1}{5^{k_i}}= b_i\frac{a_{i}^{2018}-1}{a_i +1} + b_{i+1} \equiv 2b_i + b_{i+1} \pmod 5 \implies 2b_i + b_{i+1} \equiv 0 \pmod 5$$$$b_{i+1} \equiv -2b_i \pmod 5 \implies b_{i+2} \equiv 4b_i \pmod 5 \implies b_{i+4} \equiv b_i \pmod 5$$$$b_1 \equiv b_{2019} \pmod 5 \implies b_1 \equiv b_{2019-2016}=b_3 \equiv 4b_1 \pmod 5 \implies b_1 \equiv 0 \pmod 5$$ This is contradiction because $a_1 +1 = 5^{k_1}.b_1$ with $v_5(a_1)=k_1$, this implies the problem. $\blacksquare$

I am sorry,I am not smart.Also I can finally start bashing AIME'S now yayy! India TST 2019 D1 P2 wrote: Show that there do not exist natural numbes $a_1, a_2, \cdots, a_{2018}$ such that the numbers \[ (a_1)^{2018}+a_2, (a_2)^{2018}+a_3, \cdots, (a_{2018})^{2018}+a_1 \]are all powers of $5$ Proposed by Tejaswi Navilarekallu Total ISL 2014 N5 Vibes.We proceed by contradiction assuming the opposite of the given statement. If $5|a_1$ we see $5|a_i$ for all $i$(Just assuming randomly,you can see that choosing any $i$ will work).Since $a_1^{2018}+a_2=5^k$ we get $v_5(a_1^{2018})=2018v_5(a_1)=v_5(a_2)$,so repeating this process on other terms continuously gives $2018^{2018}v_5(a_1)=v_5(a_1) \implies v_5(a_1)=0$, which is impossible. So, $$5 \not | a_1 \implies a_2 \equiv -a_1^2 \pmod 5 \implies a_3 \equiv -a_2^2 \equiv -(-a_1^2)^2 \equiv -a_1^4 \equiv -1 \pmod 5$$Practicing this we get $a_4 \equiv -1 \pmod 5$,then $a_5 \equiv -1 \pmod 5,a_6\equiv -1 \pmod 5$ and so on.....finally we derive $a_{2018} \equiv -1 \pmod 5$ which immediately implies $a_1 \equiv -1 \pmod 5 \implies a_2 \equiv -1 \pmod 5$ so we get that $5|a_i+1$ for all $i$. So we set that $a_1^{2018}+a_2=5^k$ as the minnimum of all possible $a_i^{2018}+a_{i+1}$.So $5^k|a_i^{2018}+a_{i+1}$ for all $i$. Now see that, $$a_1 \equiv -a_{2018}^{2018} \equiv -a_{2017}^{2018^2} \equiv -a_{2016}^{2018^3} \equiv \cdots -a_1^{2018^{2018}} \pmod{5^k}^*$$Thus we get $5^k|a_1^{2018^{2018}}+a_1 \implies 5^k|a_1^{2018^{2018}-1}+1$ Finally by Lifting the Exponent Lemma, $$v_5(a_1^{2018^{2018}-1}+1)=v_5(a_1+1)+v_5(2018^{2018}-1)=v_5(a_1+1) \ge v_5(5^k)=k$$But $a_1+1 < a_1^{2018}+a_2=5^k \implies \text{Contradiction !}$,thus there doesnt exist such numbers $\blacksquare$ $^*$For someone interested in how this works, $$a_3 \equiv -a_2^{2018} \equiv -(-a_1^{2018})^{2018} =-a_1^{2018^2} \pmod{a_1^{2018}+a_2=5^k}$$

Pluto1708 wrote: Kayak wrote: Show that there do not exist natural numbes $a_1, a_2, \cdots, a_{2018}$ such that the numbers \[ (a_1)^{2018}+a_2, (a_2)^{2018}+a_3, \cdots, (a_{2018})^{2018}+a_1 \]are all powers of $5$ Proposed by Tejaswi Navilarekallu Assume to the contrary there exists such numbers. Claim : $5$ does not divide $a_i$ for any $a_i$ Proof : Clearly if $5$ divides one of $a_i$ it divides all.Now consider the number $a_k^{2018}+a_{k+1}$ where $k$ is chosen so that $\nu_5(a_k)=\max{\nu_5(a_i)}$. Clearly $\nu_5(a_k^{2018}+a_{k+1})$ cant be a power of $5$ since $\nu_5(a_k^{2018}) > \nu_5(a_{k+1}) >0 $ which concludes the claim.$\square$ Next note that \begin{align*} a_1^{2018}\equiv -a_2\pmod 5 \\ a_2^{2018}\equiv -a_3\pmod 5 \\ \cdots a_{2018}^{2018}\equiv -a_1\pmod 5 \end{align*}Substituting repeatedly we get $a_1^{2018^{2018}}\equiv -a_1\pmod 5$.Thus $a_1^{2(2018^{2018}-1)}\equiv 1 \pmod 5$.Thus $5-1\mid 2(2018^{2018}-1)$.But this is a contradiction.$\blacksquare$ why $5-1\mid 2(2018^{2018}-1)$.

TFIRSTMGMEDALIST wrote: Pluto1708 wrote: Kayak wrote: Show that there do not exist natural numbes $a_1, a_2, \cdots, a_{2018}$ such that the numbers \[ (a_1)^{2018}+a_2, (a_2)^{2018}+a_3, \cdots, (a_{2018})^{2018}+a_1 \]are all powers of $5$ Proposed by Tejaswi Navilarekallu Assume to the contrary there exists such numbers. Claim : $5$ does not divide $a_i$ for any $a_i$ Proof : Clearly if $5$ divides one of $a_i$ it divides all.Now consider the number $a_k^{2018}+a_{k+1}$ where $k$ is chosen so that $\nu_5(a_k)=\max{\nu_5(a_i)}$. Clearly $\nu_5(a_k^{2018}+a_{k+1})$ cant be a power of $5$ since $\nu_5(a_k^{2018}) > \nu_5(a_{k+1}) >0 $ which concludes the claim.$\square$ Next note that \begin{align*} a_1^{2018}\equiv -a_2\pmod 5 \\ a_2^{2018}\equiv -a_3\pmod 5 \\ \cdots a_{2018}^{2018}\equiv -a_1\pmod 5 \end{align*}Substituting repeatedly we get $a_1^{2018^{2018}}\equiv -a_1\pmod 5$.Thus $a_1^{2(2018^{2018}-1)}\equiv 1 \pmod 5$.Thus $5-1\mid 2(2018^{2018}-1)$.But this is a contradiction.$\blacksquare$ why $5-1\mid 2(2018^{2018}-1)$. HELPPPPPPPPPPP PLEASE GUYS PLEASEEE

@above its incorrect, ai could be 4 mod 5

So the sol of pluto is incorrect right?

TFIRSTMGMEDALIST wrote: So the sol of pluto is incorrect right? Well, as we can see in #4 one can show that $a_1 \equiv -1 \pmod 5 \implies \text{ord}_5 (a_1) = 2$, so yeah, the solution is wrong.

This solution will use the fact that $a>0$ is not a power of $5$ if there exists $a>b>0$ such that $\nu_5(b) \geq \nu_5(a)$. Suppose otherwise; we first narrow down the value of $a_i \pmod{5}$. Claim: We cannot have $5 \mid a_i$ for any $i$. Proof: If $5 \mid a_i$ for any $i$ then $5 \mid a_i$ for all $i$. Thus pick $n$ such that $\nu_5(a_n)$ is minimal. We have $$\nu_5(a_{n-1}^{2018}+a_n)=\nu_5(a_n),$$but then $a_{n-1}^{2018}+a_n$ cannot be a power of $5$. $\blacksquare$ Thus because $2018$ is even we have $a_i^{2018} \equiv 1,4 \pmod{5}$ for all $i$, which means that we must have $a_i \equiv 1,4 \pmod{5}$. But then this means $a_i^{2018} \equiv 1 \pmod{5}$, so $a_i \equiv 4 \pmod{5}$ for all $i$. Now let $5^k$ be the least power of $5$ present in the sequence, and WLOG let $a_1^{2018}+a_2=5^k$. We have $$-a_1 \equiv a_2^{2018} \equiv a_3^{2018^2} \equiv \cdots \equiv a_{2018}^{2018^{2017}} \equiv a_1^{2018^{2018}} \pmod{5^k},$$so $5^k \mid a_1^{2018^{2018}}+a_1 \implies \nu_5(a_1^{2018^{2018}-1}+1) \geq k$ as $5 \nmid a_1$. Since $2018^{2018} \equiv 3^{2018} \equiv (-1)^{1009} \equiv -1 \pmod{5}$, by Lifting the Exponent $$k \leq \nu_5(a_1^{2018^{2018}-1}+1)=\nu_5(2018^{2018}-1)+\nu_5(a_1+1)=\nu_5(a_1+1),$$but since we must have $a_1+1<a_1^{2018}+a_2$ (otherwise $a_1=a_2=1$ which is impossible), it follows that $a_1^{2018}+a_2$ is not a power of $5$—contradiction. $\blacksquare$

Nothing new in the solution, but I will still post : P Suppose there exists such $a_1,\dots,a_n$. Let the minimum out of all powers (WLOG) be $(a_1)^{2018}+a_2=5^{\alpha}$. Since it is the minimum, we have $$ (a_1)^{2018}\equiv -a_2\mod 5^{\alpha}$$$$ (a_1)^{2018^2}\equiv -a_3\mod 5^{\alpha}$$$$\vdots$$$$ (a_1)^{2018^{2017}}\equiv -a_{2018}\mod 5^{\alpha}$$$$ (a_1)^{2018^{2018}}\equiv -a_1\mod 5^{\alpha}$$$$\implies a_1((a_1)^{2018^{2018}-1}+1)\equiv 0\mod 5^{\alpha}.$$ Claim: $5\nmid a_i\forall i\in [2018]$ Proof: If $5|a_k\implies 5|a_{k+1}\implies 5|a_{k+2}\implies \dots \implies 5|a_{k-1}$ WLOG $v_5(a_1)$ be the lowest. Then $$v_5((a_{2018})^{2018}+a_1)=v_5(a_1).$$Which is not possible. So by our, above claim, $$a_1((a_1)^{2018^{2018}-1}+1)\equiv 0\mod 5^{\alpha}\implies a_1\equiv 4\mod 5\implies a_i\equiv 4\mod 5\forall i\in [2018].$$ So we have $$(a_1)^{2018^{2018}-1}+1\equiv 0\mod 5^{\alpha}\implies (a_1)^{2018}+a_2|(a_1)^{2018^{2018}-1}+1.$$ Now, by LTE, $$v_5((a_1)^{2018^{2018}-1}+1)=v_5(a_1+1)+v_5(2018^{2018}-1)=v_5(a_1+1).$$ Since $(a_1)^{2018}+a_2$ is a power of $5$, we get that $$(a_1)^{2018}+a_2|a_1+1.$$But $(a_1)^{2018}+a_2>a_1+1$ as $a_i\equiv 4\mod 5$.

Kayak wrote: Show that there do not exist natural numbes $a_1, a_2, \cdots, a_{2018}$ such that the numbers \[ (a_1)^{2018}+a_2, (a_2)^{2018}+a_3, \cdots, (a_{2018})^{2018}+a_1 \]are all powers of $5$ Proposed by Tejaswi Navilarekallu Clearly $\nu_5(a_1)=2018\nu_5(a_{2018})=\cdots =2018^{2018}\nu_5(a_1)\implies\nu_5(a_1)=0\implies 5\nmid a_i$ for all $i=1,\cdots,2018$. Let $n\geq 1$ be the lowest power of $5$. We have $$a_1\equiv -a_{2018}^{2018}\equiv \cdots\equiv -a_1^{2018^{2018}}\pmod{5^n}\implies a_1^{2018^{2018}-1}\equiv -1\pmod{5^n}.$$Let $h=\text{ord}_5(a_1)$. Then $h\mid 4\cdot 5^{n-1}$ and $h\mid 2018^{2018}-1$ imply $h=2$ or $a_1\equiv -1\pmod{5^n}$. Hence $a_i\equiv -1\pmod{5^n}$ for all $i=1,\cdots,2018$, which is a contradiction since $(5^n-1)^{2018}+5^n-1>5^n$.

handwritten solution images attached

Latexing old unsolved problems on phone