Task 3. Find all solutions to the equation $x^{2}-[x]=1$. Here $[x]$ is the integer part of $x$, that is, the greatest integer not exceeding the given number. For example, $[2.9]=2 ;[-2.9]=-3$.
Solution. 1st method. Let $[x]=n$. Then the equation reduces to the form $x^{2}=n+1$ with the additional constraint $n \leq x<n+1$. We know the value of $x^{2}$, so it is desirable to square the inequality. This is possible if its terms are non-negative.

Since $x^{2}$ is non-negative, $n+1 \geq 0 ; n \geq-1$. The only negative value of $n$ is -1. In this case, $x=0,[x] \neq n$. There is no solution.

If $n \geq 0$, then $x$ is also non-negative. Then the inequality $n \leq x$ can be squared, yielding $n^{2} \leq x^{2}=n+1$. Solving the inequality $n^{2} \leq n+1$ we get $\left(n-\frac{1}{2}\right)^{2} \leq \frac{3}{4}$; that is, $0 \leq n \leq$ $\frac{1+\sqrt{3}}{2} \approx 1.37$. Therefore, only $n \leq 1$ fits.

If $n=0$, then $x=1,[x] \neq n$.

If $n=1$, then $x=\sqrt{2},[x]=1=n-$ solution.

2nd method. Introduce the notation $x=[x]+t, 0 \leq t<1$. Then $[x]=x-t$, the equation takes the form $x^{2}-x=1-t$. In particular, $0<x^{2}-x \leq 1$.

We can solve this system of inequalities, for example, as follows: $1 \leq 4 x^{2}-4 x+1<5 ; 1 \leq(2 x-1)^{2}<5$. Thus, $\sqrt{5} \leq 2 x-1<-1$ or $1 \leq 2 x-1<\sqrt{5}$.