[Nitin Prasad]

Given eqn-
$$x^2-pq x+(p+q)=0$$
Consider discriminant of the above equation
$$D=(pq)^2-(p+q)^2=-(p^2+q^2+pq)=-(p+\frac{q}{2})^2-\frac{3q^2}{4}\leq 0$$

$$Hence D\leq 0$$

But for roots to be real, D>=0 and hence  D=0 which is possible only when p=q=0

 

Hence no pair of natural number (p,q) exists which satisfy the given condition.

[Nitin Prasad]

Yeah, your answer is correct again.

[Nitin Prasad]

Yeah your answer is correct!

[Nitin Prasad]

Let-

1. $$A=\{ n : n<10,000 \& n\in \mathbb{N}\}$$
2. $$A_3=\{ n :n \in A \& 3|n\}$$
3. $$A_5=\{ n :n \in A \& 5|n\}$$

Hence, number of positive integers less than or equal to 10,000 which are divisible by neither 3 nor 5

$$=|A_3^c\bigcap A_5^c|$$

$$=|(A_3\bigcup A_5)^c|$$

$$=10,000-|(A_3\bigcup A_5)|$$

$$=10,000-|(A_3|-|A_5)|+|(A_3\bigcap A_5)|$$

$$=10000-\lfloor\frac{10000}{3}\rfloor-\lfloor\frac{10000}{5}\rfloor+\lfloor\frac{10000}{15}\rfloor=10000-3333-2000+666=5333$$

Yeah, your answer is correct!

[Nitin Prasad]

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 ⌋. Now observe that-
$$\lim_{n\longrightarrow\infty}\frac{⌊a⌊an⌋⌋ + n}{4⌊an⌋}=\lim_{n\longrightarrow\infty}$$

$$\Longrightarrow \frac{a^2+1}{4a}=1$$

$$\Longrightarrow a=2+\sqrt{3}\; or\; a=2-\sqrt{3}$$

a) $$ P(x)=(2-\sqrt{3})x$$ cannot be a solution for it fails to satisfy the given condition for n=1.
b) Surprisingly $$latex P(x)=(2+\sqrt{3})x$$ does satisfy our solution. The equation
$$⌊(2+\sqrt(3))⌊(2+\sqrt(3))n⌋⌋ + n = 4⌊(2+\sqrt(3))n ⌋$$ reduces to

$$⌊(2-\sqrt(3)){\sqrt(3n)} ⌋=0$$ which is indeed true. Calculation part for this shown below

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