4. Determine all pairs of integers $(a, b)$ that solve the equation $a^{3}+b^{3}+3 a b=1$.
SOLUTION: The equation is symmetric in $a$ and $b$, so we can limit ourselves to considering solutions with $a \geq b$. There are no solutions with both $a$ and $b$ positive: indeed, in such a case we would have $a^{3}+b^{3}+3 a b \geq 1+1+3=5>1$, which is absurd. If at least one of $a$ and $b$ is equal to zero, say $b$, then substituting in the text we find $a^{3}=1$, i.e., $a=1$, leading to the two solutions $(1,0)$ and $(0,1)$. If then $a$ and $b$ are both strictly less than 0, then (still assuming $b \leq a < 0$ and $bc$, while from the first we find $3 a c+1=(a-c)\left(c^{2}+c a+a^{2}\right)$. Since $a-c \geq 1$ we then have

$$
3 a c+1=(a-c)\left(c^{2}+c a+a^{2}\right) \geq c^{2}+c a+a^{2},
$$

from which $1 \geq c^{2}-2 c a+a^{2}$, or $(a-c)^{2} \leq 1$. Since we have already observed that $a-c$ is positive, we must have $a-c=1$, and substituting $a=c+1$ in (??) we find

$$
c^{3}+3 c^{2}+3 c+1=c^{3}+3 c(c+1)+1,
$$

an equation that is satisfied for any value of $c$. Remembering that at the beginning we had assumed $a \geq b$, we then see that the solutions of the proposed equation are the pair $(a, b)=(-1,-1)$ and the infinite pairs $(a, b)=(a,-c)=(c+1,-c)$ for any integer value of $c$: those with $c \geq 0$ are those for which $a \geq b$, and those with $c<0$ are the symmetric ones with $a<b$.

SECOND SOLUTION: We observe that

$$
a^{3}+b^{3}-1+3 a b=a^{3}+b^{3}+(-1)^{3}-3 a b \cdot(-1)=(a+b-1)\left(a^{2}+b^{2}+1-a b+a+b\right),
$$

so pairs of the form $(n, 1-n)$ are solutions for any integer $n$.

Multiplying the second factor by 4, we can write

$$
4 a^{2}+4 b^{2}+4-4 a b+4 a+4 b=(2 a-b+1)^{2}+3(b+1)^{2},
$$

which, being a sum of squares, vanishes only when $(2 a-b+1)=(b+1)=0$, i.e., when $a=$ $b=-1$.