Here is a sketch of my solution: A lemma: Let $S$ be a set of infinitely many polynomials such that the product of any two elements of $S$ are in $Q[x]$ then the polynomials are of the form: $P(x)= cR(x)$ where $R(x)$ a rational coefficient polynomial and $c$ is a number (note that c may be different for different polynomials) Proof: First, divide all rational polynomials out of $P(x)$ (easy to see that $S$ still works) now assume the contrary, there still exists a nonconstant polynomial. Call it $R(x)$. $R(a)=0$ also let $Q$ and $H$also be elements of $S$. $a$ is a root of the rational polynomial $QR$ so ther exists polynomial $T$ such that $T$ is the minimal polynomial of $a$. Let $T=T_1 T_2$ such that $T1$ divides $R$ and $T_2$ and $R$ share no roots. Looking at $QR$ and $HR$ We conclude that $T_2$ divides $Q$ and $H$ now looking at $QH$ we get that $T_2$ divides $QH$ now continuing this we get that $T$ must divide an element of $S$, is a contradiction. So all remaining polynomials are constant which means their nonconstant parts where rational. We return to the problem at hand, interesting polynomials are obviously rational so according to the lemma we have: $P_i(x)=Q_i(x) c_i$ Now notice that the leading coefficient of $P_i(x)$ is $1$ or $-1$ Now it is easy to see that $P_i(1)=\sqrt{1399} or -\sqrt{1399}$ so $c_i=q\sqrt{1399}$ where $q$ is nonzero and rational. Now: $P_i(x)=Q_i(x)q\sqrt{1399}$ Comparing the leading coefficient of both sides, one side is rational and the other irrational(because $1399$ is not a square) a contradiction so no such polynomials exist.

Maybe another proof for the lemma above: Note that $P_1(x)^2=\frac{P_1(x)P_2(x) \cdot P_1(x)P_3(x)}{P_2(x)P_3(x)},$ Both sides are polynomials, so comparing the coefficients we get that $P_1(x)^2 \in \mathbb Q[x].$ Let $P_1(x)^2=qx^{2n}+\cdots,$ Again by comparing coefficients, we get that all coefficients of $P_1(x)$ are in $\sqrt{q} \mathbb Q$. This proves the lemma. I wonder if the proof is correct, since in this way we only use $3$ polynomials to get the contradiction.

I solved the problem by $2$ ways in the exam. The first one is noted above. here is the second one. take The biggest polynomial (in terms of degree) like $Z(x)$ with $deg(Z)=D_Z$. such that we can write each $P_i(x)$ as $Z(x)+r_i x^{D_z+c_i}+x^{D_z+c_i+1}(Q_i(x))$. where $c_i>0$. now just solve for when $c_i$ is not bounded (which is easy). and for the bounded part make $r_i$ have some finite choices for it. Then as we have infinitly many of them. choose the Polynomials which all have the same $r_i$ and $c_i$. now our $Z(x)$ became bigger. from here the problem is not hard to finish.

acctualy there isn’t even 3 of them !!! take for example $p_1(x),p_2(x),p_3(x)$ first I’ll prove with induction that the coefficients of these three polynomials are rational. first it’s easy to see the fixed term of this three polynomials are either 1 or -1 now assume we know the coefficients of these polynomials are rational up to $x^k$ now I’ll prove the the coefficient of $x^{k+1}$ in all of these polynomials is rational assume the coefficients of $x^{k+1}$ in $p_1(x)$ is $a_1$ and in $p_2(x)$ is $a_2$ and ... now by checking the coefficients of $x^{k+1}$ in $p_1(x)p_2(x)$ and using the induction hypothesis we will find out that $a_1+a_2$ is rational and doing the exact same thing for $p_2(x)$ and $p_3(x)$ we will find out $a_2+a_3$ is rational as well as $a_1+a_3$ . so all of them are indeed rational(cause $a_2+a_3-(a_1+a_3)$ is rational) so we proved the coefficients of these polynomials are rational but we know $p_1(1)p_2(1)$=1399 and $p_1(1)p_3(1)$=1399 and $p_2(1)p_3(1)$=1399 so $p_1(1)$=$\sqrt{1399}$ but the coefficients of $p_1(x)$ was rational contradiction. so we are done.

@above , yes , the advantage of the first solution is that it shows there isn't even $3$ of them. but i do think you might get stuck if it were for example $n^2-1$ istead of $1398$. . the second solution solves for every positive integer $n$. but on the other hand , it actually uses that there are infinitly many of them. . now as a new problem , if instead of $1398$ we had $n^2-1$ could you prove there arn't $3$ polynomials that satisfy the problem? i can do it for "finitly" many , but not for an specific number like $3$ or $4$. i would be intrested if anyone could solve these as well

I may have misunderstood your problem but: For $k^2 -1$ doesn’t $P_i (x) = x^{k^i (k-1)} +x^{k^i (k-2)} + \dots + x^{k^i} +1$ work?

seems like i have made an assertion on$Z$ and $D_Z$ which is not true in the case of $n^2-1$. thanks for correcting me. but with those extra assumptions it seems that i can solve for $n^2-1$ aswell. which is not the problem anymore. its some thing with more assumptions

Mr.C wrote: We call a polynomial $P(x)$ intresting if there are $1398$ distinct positive integers $n_1,...,n_{1398}$ such that $$P(x)=\sum_{}{x^{n_i}}+1$$Does there exist infinitly many polynomials $P_1(x),P_2(x),...$ such that for each distinct $i,j$ the polynomial $P_i(x)P_j(x)$ is interesting. Were the polynomials meant to be distinct? If not, suppose that each $P_i(x)$ is equal to some $S(x) \in \mathbb R [x]$. Then we want to show that $$ S(x)^2 $$cannot be interesting. This seems intuitively true, but I am not able to find proof for it. So can someone please clarify?