Do there exist a sequence $a_{1}, a_{2}, a_{3}, \ldots$ of real numbers and a non-constant polynomial $P(x)$ such that $a_{m}+a_{n}=P(mn)$ for every positive integral $m$ and $n?$
Proposed by A. Golovanov
mathmanman wrote:
Do there exist a sequence $a_{1}, a_{2}, a_{3}, \ldots$ of real numbers and a non-constant polynomial $P(x)$ such that $a_{m}+a_{n}=P(mn)$ for every positive integral $m$ and $n?$
It does not :
$m=1$ $\Rightarrow $ $a_{n}=P(n)-a_{1}$
So we have $P(m)+P(n)-2a_{1}=P(mn)$ and for example $2P(n)-2a_{1}=P(n^{2})$ which can't be true $\forall n$ (take $n\rightarrow+\infty$) if $P(x)$ is non constant.
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Patrick
mathmanman wrote:
Do there exist a sequence $a_{1}, a_{2}, a_{3}, \ldots$ of real numbers and a non-constant polynomial $P(x)$ such that $a_{m}+a_{n}=P(mn)$ for every positive integral $m$ and $n?$
We get $a_{n}=P(n)-a_{1}, \ a_{1}=P(1)/2.$ If $deg(P)=k$, then for big n $|a_{k}|n^{k}/2=c_{1}n^{k}<|P(n)|<2a_{k}n^{k}=c_{2}n^{k}$.
Therefore $|P(n^{2})|>c_{1}n^{2k}>2c_{2}n^{k}>|P(n)|+|P(n)|$ - contradition.