P.S. I didn't appear for this Olympiad, this is just a spontaneous solution.

Was this from Day 1 or day 2? And, Juniors or seniors? Please specify in source the day, problem number and category, to avoid confusion while adding this to contest collections

Math-wiz wrote:

P.S. I didn't appear for this Olympiad, this is just a spontaneous solution. It's day 1 for Juniors problem 3

Math-wiz wrote:

P.S. I didn't appear for this Olympiad, this is just a spontaneous solution. why gcd(n+1,n+2)=1 makes a_n not a perfect power?

mtahh wrote: Math-wiz wrote:

P.S. I didn't appear for this Olympiad, this is just a spontaneous solution. why gcd(n+1,n+2)=1 makes a_n not a perfect power? Because if $a_n=x^y$ for positive integers $x,y$, then $y|n+1$ and $y|n+2$ which is not possible as $gcd(n+1,n+2)=1$

if i'm not mistaken, you have skipped the most important part in this problem, which is proving this math-wiz said wrote: One can prove inductively that for $n\geq 2$, $\nu_2(b_n)=n+1$ and $\nu_3(b_n)=n$

It can be done easily by induction. Base case $k=2$ is true $b_2=6b_1(b_1+1)=6*3*4=2^3*3^2$ $\nu_2(b_2)=3=2+1, \nu_3(b_2)=2$ For $k=3,4,\ldots n$, let $\nu_2(b_k)=k+1, \nu_3(b_k)=k$ We prove it for $k=n+1$ $\nu_2(b_{n+1})=\nu_2(6b_n(b_n+1))=\nu_2(6)+\nu_2(b_n)+\nu_2(b_n+1)=1+(n+1)+0=n+2$ $\nu_3(b_{n+1})=\nu_3(6b_n(b_n+1))=\nu_3(6)+\nu_3(b_n)+\nu_3(b_n+1)=1+n+0=n+1$ And the induction is complete

A nice generalisation for this one could be- For what initial positive integral values of $a_1$ does the sequence $a_n=a_{n-1}^2+6a_{n-1}, n>1$ have no perfect powers? Any ideas for this one? Here are few more possibilities of generalisations, but I have no clue if they are solvable. For what initial values does the sequence have atleast one perfect power? For what initial values does the sequence have infinitely many perfect powers?

Assume that for some positive integer $n_0$ and some prime $p>3$ we have $v_p(a_{n_0})=1$ then we will prove that for all $n \ge n_0$ we have $v_p(a_n)=1$.The proof is of course inductive since $a_{n-1}|a_{n}$ for all $n$ all the terms after $a_{n_0}$ will be divisible by $p$ and: $v_p(a_{n+1})=v_p(a_n)+v_p(a_n+6)=v_p(a_n)=1$.The claim shows that if we find such $p,n_0$ we will be done and: $a_1=18,a_2=432,a_3=432*438$ so we choose $n_0=3,p=73$ the first terms can easily be checked by bare hands.

It is easy to compute $a_2 = 432$ so $v_2(a_2) = 4$ and $v_3(a_2) = 3$. Since $a_n = a_{n-1}(a_{n-1} + 6)$, we have $v_2(a_n) = v_2(a_{n-1}) + v_2(a_{n-1} + 6) = 1 + v_2(a_{n-1})$ for all $n\geq 3$. And so we have $v_2(a_n) = n + 2$ for all $n\geq 3$. By a similar argument for $v_3$, we have $v_3(a_n) = v_3(a_{n-1}) + v_3(a_{n-1} + 6) = 1 + v_3(a_{n-1})$ for all $n\geq 3$. So we have $v_3(a_n) = n + 1$ for all $n\geq 3$. So for $n\geq 3$, since $v_2(a_n) = n + 2$ and $v_3(a_n) = n + 1$ and $\gcd{(n + 2 , n + 1)} = 1$, we cannot $a_n$ to be a perfect power and it is easy to see that $a_1$ and $a_2$ are both non-perfect powers. So we are done. $\square$

bigant146 wrote: Let $a_n$ be a sequence given by $a_1 = 18$, and $a_n = a_{n-1}^2+6a_{n-1}$, for $n>1$. Prove that this sequence contains no perfect powers. $a_2 = a_1(a_1+6) = 18 \cdot 24 = 432,a_3 = a_2^2+6a_2 = 432 \cdot 438$ $$1 = v_{73}({a_3}) = v_{73}({a_4}) = ... = v_{73}({a_n}) = ...$$Remaining easy.$\blacksquare$

Close it at consecutive square