Is there a polynomial $P(n)$ such that $nP(n) = \varphi(n)$, or also $\varphi(n)P(n)$ is a square number?

Question

1. is there a polynomial with real coefficients $P$ such that $nP(n) = \varphi(n)$ for all $n \ge 1$?

2. is there a non-constant polynomial with integer coefficients $P$ such that $P(n)\varphi(n)$ is a square number for all $n \ge 1$?

where $\varphi(n)$ is the number of natural numbers less than or equal to $n$ that is relative prime to $n$.

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Attempt:

For no. 1.), let us see if $n$ is prime, let say $n=x$ is prime, then always $\varphi(x) = x-1$. So if $nP(n) = \varphi(n)$ then $$ P(x) = \frac{x-1}{x} $$

Now notice that $0 < \frac{x-1}{x} < 1$. But since any polynomial has limit $\pm \infty$ as $x \rightarrow \infty$, and prime number is unbounded above, so there must exist a large enough prime for any polynomial such that $P(x)>1$. So the answer is there is no such polynomial.

For no. 2), I have difficulties..

Answer 1

For question 2., let's assume such a polynomial

$$P(x)=p_mx^m+\ldots+p_1x+p_0$$

exists, and that it has among all such polynomials the minimal degree $m$.

There are 2 cases, regarding the constant term $p_0$:

Case 1: $p_0 \neq 0$

Then there is a prime number $q_1$ with $q_1 \nmid p_0$, from which follows that $q_1 \nmid P(q_1^k)$ for any $k \ge 1$. So we have

$$P(q_1^2)\varphi(q_1^2) = P(q_1^2)q_1(q_1-1)$$

and the right hand side contains the prime $q_1$ in exactly the 1st power. So it cannot be a square!

Case 2: $p_0=0$

We have $P(x)=xR(x)$ with

$$R(x)=r_{m-1}x^{m-1}+\ldots+r_1x+r_0$$

again being a polynomial with integer coefficients. Again we need to consider 2 cases:

Case 2.1: $r_0 \neq 0$

Again we find a prime $q_2 \nmid r_0 \Longrightarrow q_2 \nmid R(q_2)$ and get

$$P(q_2)\varphi(q_2)=q_2R(q_2)(q_2-1)$$

with the right hand side containg $q_2$ in the 1st power, so not a square.

Case 2.2: $r_0 = 0$

Again we get $P(x)=x^2S(x)$, with $R(x)=xS(x)$, with $S(x)$ having integer coefficients.

So for any $n \ge 1$ we have

$$ \mathbb Z \ni S(n)\varphi(n) = \frac{P(n)}{n^2}\varphi(n) = \frac{P(n)\varphi(n)}{n^2}$$

and the rightmost term is the square of rational number, so a square of an integer as it is itself an integer.

So $S(x)$ is also a polynomial that has the properies required of problem 2. But since $P(x)$ is non-constant, $S(x)$ has a smaller degree than $P(x)$, which contradicts the choice of $P(x)$.

That proves that such a polynomial cannot exist at all.

Answer 2

Alternative answer for 1:

If $nP(n) = \varphi(n)$ for all $n\ge1$, then $xP(x)-(x-1)$ is a polynomial that has an infinite number of zeros (the prime numbers). Therefore, $xP(x)-(x-1)$ is the zero polynomial. But then $nP(n) \ne \varphi(n)$ when $n$ is not prime.