Obviously $x=0$ satisfies the given equation. Also, $x=n$, where $n$ is a positive integer, does not satisfy the given equation, because $n(n^2+1)>n^3.$ We need to show that there is a solution for any interval $(n, n+1).$ Indeed, for $x=n+t,$ where $0<t<1$, we have $n(n^2+2nt+t^2+1)>n^3$ and $n(n^2+2nt+t^2+1)<(n+1)^3\leftrightarrow n^2(3-2t)+n(2-t^2)+1>0,$ which is true because $\Delta<0$ for $0<t<1.$ Also, the equation $n(n^2+2nt+t^2+1)=(n+t)^3\leftrightarrow f(t)=t^3+2nt^2+n^2t-n=0\ (1)$ has a unique real solution because $f(0)\cdot f(1)<0$ and $f'>0$ in $0<t<1.$ For the second part, set $t=p/q$ in $(1)$ where $p,q$ are positive integers with $q\ne 1$ and $(p,q)=1.$ It is eaasy to see that $p^3/n$ and then $q/1$ which gives us a contradiction.

1. suppose $a \in N$ we need to show that there is a solution in $(a,a+1)$ we take $f(x)=x^3-a(x^2+1)$. if we show that there is a root of $f(x)$ in $(a,a+1)$ , then we are done. now , $f(x)$ is continuous in $[a,a+1)$ now , it's trivial to see that ,$f(a)=-a<0$ and $f(a+\frac{3}{4})>0$ so , by IVT , we get a root of $f(x)$ in $(a,a+\frac{3}{4})$ now , this root definitely satisfies $x^3=[x](x^2+1)$.....(*) clearly , any root of (*) in $(a,a+1)$ is a root of $f(x)$ and any root of $f(x)$ in $(a,a+1)$ is a root of $x^3=[x](x^2+1)$ now , note that , $f(x)$ is increasing [strictly] in $[a,a+1)$. so, there exists only one root in $(a,a+1)$ so , there exists exactly one root of $x^3=[x](x^2+1)$ in $(a,a+1)$ for any $a$ 2.note that , no positive integer satisfies the master equation. if possible ,suppose it has a positive rational root $r \in (k,k+1)$ for some nonnegative integer $k$. then $r$ satisfies $x^3-k(x^2+1)=0$ ,which is a monic polynomial. so $r$ must be an integer [positive].we have already noted that no positive integer satisfies the equation.so , a contradiction!!

Let $\lfloor x \rfloor =m$ and $\{x\} = k.$ Since $x=m+k,$ we substitute it in and find that $$m(m+k)^2+1=(m+k)^3 \implies$$$$k^3+2mk+mk-m=0.$$ Let the above polynomial be $f(k).$ Notice that $f(0)=-m<0$ and $f(1)=m^2+m+1 >0.$ This shows that there is a root for $k \in (0,1).$ Now, we show there is only one root. Suppose $r$ is a root. For all $k>r$, we have $f(k)>f(r).$ Similarly, for all $k<r$, we have $f(k)<f(r).$ This shows that there is only one root. Since there is only one value of $k$ satisfying each value of $\lfloor x \rfloor$, we have a unique solution between each consecutive integer.

Why does for $k > r$, $f(k) > f(r)$, and vice versa imply there is one root?

Jay17 wrote: Why does for $k > r$, $f(k) > f(r)$, and vice versa imply there is one root? Monotonicity

Williamgolly wrote: Jay17 wrote: Why does for $k > r$, $f(k) > f(r)$, and vice versa imply there is one root? Monotonicity If the cubic had a double root wouldn't it be monotonic, yet have more than one root? @below wait I think i understand now, sorry

I cant understand what you mean exactly, but m is positive.

nsato wrote: (a) Show that the equation \[\lfloor x \rfloor (x^2 + 1) = x^3,\]where $\lfloor x \rfloor$ denotes the largest integer not larger than $x$, has exactly one real solution in each interval between consecutive positive integers. (b) Show that none of the positive real solutions of this equation is rational. As $x=[x]+\{x\}$ so $[x](x^2+1)=x^3\implies \{x\}^3+2*[x]*\{x\}^2+\{x\}*[x]-[x]=0=p(\{x\})$ also $[x]\in N, \{x\}\in (0,1)$ Now observe $p'(\{x\})>0$ for $\{x\}\in (0, 1)$ so $p(\{x\})$ is increasing. And also $p(0)<0$ and $p(1)>0$ so $p(\{x\})$ must have only one real root in $(0, 1)$ for every $[x]\in N$ hence "a" Is proved!! Now let $\{x\}=\frac{p}{q}$ be the root of $p\{x\}$ for some $q>p, \{p, q\}\in N$ and $gcd(p, q)=1$. Then We have $[x]*q^3=p(p^2+2*[x]*p+[x]*q^2)$ So $p|[x]\implies [x]=p*k$ but then $k|(p^2+2*[x]*p+[x]*q^2)$ which is not possible.. Hence "b" Is proved.

Let $\lfloor x\rfloor = k$. We fix $k$ and want a solution between $k$ and $k+1$. Rearranging the equation gives $x^3-kx^2-k = 0$. Substitute $k$ into the equation to get $-k$. Substitute $k+1$ into the equation to get $k^2+k+1$. There must be a root from $k$ to $k+1$ because $-k$ is negative and $k^2+k+1$ is positive for all positive integers $k$. Note that if we take the derivative of $x^3-kx^2-k=0$, we get $3x^2-2kx$. The roots of the derivative are $x = 0$ and $x = \frac{2k}{3} > k$. Thus, there are no turns in the graph so there is exactly $1$ solution from $k$ to $k+1$. To show that $x$ can't be rational, we let $x = \frac{m}{n}$ where $m$ and $n$ are relatively prime integers. Note that \[\lfloor x \rfloor = \frac{x^3}{x^2+1}\]. We want $\frac{x^3}{x^2+1}$ to be an integer. Substitute $x = \frac{m}{n}$ to the expression to get \[\cfrac{\frac{m^3}{n^3}}{\frac{m^2+n^2}{n^2}} = \cfrac{m^3}{(n)(m^2+n^2)}\]Note that because $m$ and $n$ are relatively prime, the $n$ in the denominator will never get canceled out. Thus, the expression will never be an integer, contradiction.