oVlad wrote: It is well-known that a quadratic equation has no more than 2 roots. Is it possible for the equation $\lfloor x^2\rfloor+px+q=0$ to have more than 100 roots? Yes. Choose for example $p=q=0$ : any $x\in(-1,1)$ is a root.

pco wrote: oVlad wrote: It is well-known that a quadratic equation has no more than 2 roots. Is it possible for the equation $\lfloor x^2\rfloor+px+q=0$ to have more than 100 roots? Yes. Choose for example $p=q=0$ : any $x\in(-1,1)$ is a root. My bad, I completely forgot to write the condition $p\neq 0$.

oVlad wrote: It is well-known that a quadratic equation has no more than 2 roots. Is it possible for the equation $\lfloor x^2\rfloor+px+q=0$ with $p\neq 0$ to have more than 100 roots? Let $x^2=n+y$ with $n\in\mathbb Z_{\ge 0}$ and $y\in[0,1)$ Equation is $n\pm p\sqrt{n+y}+q=0$ and so $p^2(n+y)=(q+n)^2$ with one root per $(n,y)$ (choosing sign of $ x$ as sign of $-\frac{q+n}p$ The constraint is $n+1>\frac{(q+n)^2}{p^2}\ge n$ and so the system $n^2+n(2q-p^2)+q^2\ge 0$ and $n^2+n(2q-p^2)+q^2-p^2<0$ Here, choosing for example $p^2=4q$, constraint becomes $(n-q)^2<4q$ which gives $\sim 4\sqrt q$ possibilities. Choosing for example $q=900$, $p=60$ and any $n\in(840,960)$, $\boxed{\text{We have 119 solutions for equation }\left\lfloor x^2\right\rfloor+60x+900=0\text{ : }x_n=-\frac{n+900}{60}\forall n\in\{841,842,...,958,959\}}$