lemma:there exist $x_{0}$ such that $q(p(x_{0}))+h(x_{0})=x_{0}$ proof:since $p(x)$ is descending we know that $p(x)$ cut the line $y=nx+m$ in exactly one point name that point $x_{0}$ so we have $p(x_{0})=nx_{0}+m$ and this point is unique.put this in the problem we get $p(q(nx_{0}+m)+h(x_{0}))=n(q(nx_{0}+m)+h(x_{0}))+m$ since $x_{0}$ is unique we have $q(nx_{0}+m)+h(x_{0})=x_{0}$ but $p(x_{0})=nx_{0}+m$ so $q(p(x_{0}))+h(x_{0})=x_{0}$ hence we are done. so if there exist a function like f we have $f(x_{0})=f(x_{0})^2+1$ which is imposible

mmaht wrote: since $p(x)$ is descending we know that $p(x)$ cut the line $y=nx+m$ in exactly one point please explain more for this part !!

alibez wrote: mmaht wrote: since $p(x)$ is descending we know that $p(x)$ cut the line $y=nx+m$ in exactly one point please explain more for this part !! $n\in \mathbb{N}\Rightarrow y=nx+m$ is ascending

alibez wrote: mmaht wrote: since $p(x)$ is descending we know that $p(x)$ cut the line $y=nx+m$ in exactly one point please explain more for this part !! p(x) is descending so the degree of p(x) is odd and we know that ${p}'(x)< 0$ (because $p(x)$ is descending) so if we assume that $l(x)=p(x)-nx-m$ we have ${l}'(x)={p}'(x)-n< 0$ so $l(x)$ is also descending so $l(x)$ is one to one.but since the degree of $p(x)$ is odd we have the degree of $l(x)$(you may ask how if p(x) is linear?then we have p(x)=ax+b and since p(x) is descending $a\leq 0$ so $a\neq n$ ) is odd too so it has one root and since l(x) is one to one this root is unique.

Hello... I have a problem and i'd be happy if anyone shows me how to solve it. It is as follows: find the greatest value of a such that a^5 + 5^5 is divisible by a-5.

In principle, do open a new thread for a different problem. But here's the answer. To have $a-5 \mid a^5+5^5$ means $a-5 \mid (a^5 - 5^5)+(5^5 + 5^5)$, so $a-5 \mid 2\cdot 5^5$. The largest such $a$ is $a=2\cdot 5^5 + 5$, and it clearly verifies.

Funny problem . Here goes my solution . (Took way longer than it should have becoz I kept misreading the brackets ) Set $g(x) \stackrel {\text {def}}{=} nx+m$ . Note that since $p(x)$ is a decreasing function and $g(x)$ is a increasing function and both are continuous, hence there exists a unique $c \in \mathbb R$ satisfying $P(c)=nc+m$ . Plugging in $c$ into the equation gives $:=$ $$p (q( \underbrace {nc+m}_{p(c)}) +h(c) ) = n(q( \underbrace {nc+m}_{p(c)})+h(c)) +m $$$$ \implies p( q(p(c))+h(c) ) = n (q(p(c))+h(c))+m $$ Hence if $d= q(p(c))+h(c)$ , then $p(d)=g(d)$ . Since $c$ was unique, we have $q(p(c))+h(c)=c$ . Assuming the problem statement to be false, we have $$f(q(p(c))+h(c))=f(c) < f(c)^2+1$$We are done .$\blacksquare$