Consider some polynomial P(x) with degree>1. Observe that for large values of natural numbers n, $$\frac{4P(n)}{P(P(n))+n}$$ is approximately 0. Using this idea it seems that our polynomial must be a constant polynomial. Though these were loose arguments and now we need to make our arguments precise. For this, few lemmas have been stated without proof.
Lemma 1 : Consider polynomial P(x) ∈ R[x] of degree n>0 such that its leading coefficient is positive. It can be shown that there exist natural number m such that P(x) is strictly increasing function in the interval [m,\infty).
Lemma 2: Consider polynomial P(x) ∈ R[x] of degree n>1 such that its leading coefficient is positive. Then there exist natural number m such that P(x)>m ∀ x>m and is strictly increasing in the interval [m,\infty)
Proof sketch: Consider polynomial Q(x)=P(x)-x. By lemma 1, there exist m such that Q(x)is strictly increasing in [m,\infty). Hence P(x)-x>0 for all x>m i.e. P(x)>x>m for all x>m.
Solution: We consider following cases for P(x) ∈ R[x]-
1) Degree of P(x)>1 & leading coefficient is positive: Then by the above lemma 2, there exist natural number m such that P(x) is strictly increasing function in the interval (m,\infty) and P(x)>m for all x>m . Hence we have-
$$P(P(n)-1))+n-1<P⌊P(n)⌋-1+n <⌊P⌊P(n)⌋⌋ + n = 4⌊P(n)⌋<4(P(n)+1) ∀n ∈ N.$$
$$\Longrightarrow P(P(n)-1))-4P(n)+n-5<0 ∀n ∈ N.$$
Now observe that P(P(n)+1))-4P(n)+n+5 is a polynomial of degree>0 with positive leading coefficient. Hence by lemma1 we arrive at a contradiction.
2) Degree of P(x)>1 & leading coefficient is negative: Similar to above case and leads to a contradiction
3) Degree of P(x) at most 1: Consider P(x)=ax+b. Since P(0)=0 Hence b=0. Now we have ⌊a⌊an⌋⌋ + n = 4⌊an ⌋
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