Let's find the polynomial $p(x)$ with integer coefficients of the lowest degree, such that its leading coefficient is 1, and it satisfies $p(0)=0$, $p(1)=1$, and $p(-1)=3$.
Solution. The conditions are: $p(0)=0, p(1)=1, p(-1)=3$,

$$
p(x)=1 \cdot x^{n}+a_{1} x^{n-1}+\ldots+a_{n-1} x+a_{n}, \quad n \geq 1, \quad n \in \mathbb{N}
$$

Substitute $x=0$ into the polynomial: we get $a_{n}=0$.

If $n=1$, then $p(x)=x+0$. For this polynomial, $p(-1) \neq 3$.

If $n=2$, then $p(x)=x^{2}+a_{1} x+0$. Substitute $x=1$: $p(1)=1+a_{1}=1$, so $a_{1}=0$. The obtained polynomial is $p(x)=x^{2}$, but $p(-1) \neq 3$.

If $n=3$, then $p(x)=x^{3}+a_{1} x^{2}+a_{2} x+0$.

Substitute $x=1$: $p(1)=1+a_{1}+a_{2}=1$, so $a_{1}+a_{2}=0$.

Substitute $x=-1$: $p(-1)=-1+a_{1}-a_{2}=3$, so $a_{1}-a_{2}=4$. From this: $a_{1}=2, a_{2}=-2$.

The sought polynomial is thus $p(x)=x^{3}+2 x^{2}-2 x$, since it satisfies all conditions.