Given are real numbers $a_1, a_2, \ldots, a_n$ ($n>3$), such that $a_k^3=a_{k+1}^2+a_{k+2}^2+a_{k+3}^2$ for all $k=1,2,...,n$. Prove that all numbers are equal.
Problem
Source: Stars of Mathematics 2022, J2
Tags: algebra
07.09.2023 01:02
any ideas?
07.09.2023 03:44
first of all, it is easy to see that $a_k$ is non-negative since $a_k^3 = a_{k+1}^2+a_{k+2}^2+a_{k+3}^2\geq 0$ $a_k \geq 0$ WLOG $\max(a_1,..,a_n) = a_t$ , $ 0 \leq t\leq n$ $a_{t+1}^2+a_{t+2}^2+a_{t+3}^2 \geq a_t^2+a_{t+1}^2+a_{t+2}^2$ $a_{t+3}^2 \geq a_t^2$ $a_{t+3} =a_t=k$ Compare $k = t+1 $ $a_{t+1} \geq a_{t+4}$ Compare $a_t^3 = a_{t+3}^3$ $a_{t+2} \geq a_{t+5}$ However, since it is an equality $a_{t+1}=a_{t+4} , a_{t+2} = a_{t+5},a_{t+3} = a_{t+6}$ Again Repeat the process like before $a_{t+3} = a_{t+6}$ we gonna get $a_{t+4} = a_{t+7} , a_{t+5} = a_{t+8}, a_{t+6} = a_{t+9}$ so each who has the same index mod 3 is equal but all the terms $a_k^3$ have the same amount of things no matter what the index is which is $a_{k+1}^2+a_{k+2}^2+a_{k+3}^2= a_0^2+a_1^2+a_2^2$ so all a_k equal $k^3 = 3k^2$ $k^2(k-3) = 0$ $a_k = 0 $ or $a_k = 3$ for all k
07.09.2023 04:53
Obviously all the $a_i$ are nonnegative, and if some $a_i$ equals zero then $a_{i+1}=a_{i+2}=a_{i+3}=0$ and by induction every term is $0$, so suppose they're all positive. If $M$ is the maximal value of the $a_i$ then we have $M^3 \leq 3M^2 \implies M \leq 3$. Similarly if $m$ is the minimal value of the $a_i$ then we have $m^3 \geq 3m^2 \implies m \geq 3$, i.e. all the $a_i$ are $3$.