1. Among all polynomials with leading coefficient 1, which give a remainder of 1 when divided by $x+1$, and a remainder of 2 when divided by $x^{2}+1$, find the one with the lowest degree.
III/1. It is clear that the degree of such a polynomial is at least 2. If such a polynomial were of degree 2, it would be of the form

$$
p(x)=x^{2}+a x+b=\left(x^{2}+1\right)+a x+(b-1)
$$

Since the remainder upon division by $x^{2}+1$ is 2, it follows that $a x+(b-1)=2$, hence $a=0$ and $b=3$. However, $p(x)=x^{2}+3=(x-1)(x+1)+4$ does not give a remainder of 1 upon division by $x+1$. This can also be seen by calculating $p(-1)=4$, since $p(-1)$ is precisely the remainder upon division of the polynomial $p(x)$ by $x+1$, which is not equal to 1.

Therefore, such a polynomial must be of degree at least 3. Let

$$
p(x)=x^{3}+a x^{2}+b x+c=(x+a)\left(x^{2}+1\right)+(b-1) x+c-a
$$

Again, it follows that $(b-1) x+c-a=2$ or $b=1, c=a+2$. Thus,

$$
p(x)=x^{3}+a x^{2}+x+a+2=(x+1)\left(x^{2}+(a-1) x+2-a\right)+2 a
$$

from which it follows that $2 a=1$ (which can also be seen from the relation $1=p(-1)=-1+a-1+a+2$). The sought polynomial is $p(x)=x^{3}+\frac{1}{2} x^{2}+x+\frac{5}{2}$.
The degree of such a polynomial is at least 2 ..... 1 point
The expression $p(x)=\left(x^{2}+1\right)+a x+(b-1)$ ..... 1 point
The expression $p(x)=x^{2}+3=(x-1)(x+1)+4$ or the calculation $p(-1)=4$ ..... 1 point
The polynomial is at least of degree 3 ..... 1 point
The expression $p(x)=(x+a)\left(x^{2}+1\right)+(b-1) x+c-a$ ..... 1 point
The expression $p(x)=(x+1)\left(x^{2}+(a-1) x+2-a\right)+2 a$ or $p(-1)=2 a$ (or an equivalent, $\mathbf{v}$
in which the numbers $b$ and $c$ may also appear) ..... 1 point
Such a polynomial is $p(x)=x^{3}+\frac{1}{2} x^{2}+x+\frac{5}{2}$ ..... 1 point