Solving $a(a + 1) + b(b + 1) = 12, (a+ 1)(b+1) = 4$, is there any trick?

Solution 1:

Let $u = a+1, v=b+1$. The equations become

$$(u-1)u+(v-1)v = 12, uv = 4$$

Thus $uv+(u-1)u+uv+(v-1)v = u(v+u-1) + v(u+v-1) = (u+v)(u+v-1) = 20$.

Let $x = u+v$ and we have a quadratic, which yields $u+v = 5$ or $-4$.

Solve the rest using this info and $uv=4$, which is just another quadratic.

EDIT: "Solving the quartic" doesn't seem too bad either, by substitution:

$$(u-1)u+\frac4u\left(\frac4u-1\right)=12$$

$$u^2+\frac{16}{u^2}-u-\frac4u=12$$

Note that $\left(u+\frac4u\right)^2=u^2+8+\frac{16}{u^2}$. Hence we can rewrite the above into:

$$\left(u+\frac4u\right)^2 - \left(u+\frac4u\right)=20$$

which, of course, is a quadratic equation with $u+\frac4u = 5$ or $-4$, and thus we arrive at essentially the same conclusion as the above (slightly unnatural) method.

Solution 2:

Expressed in elementary symmetric polynomials $s=a+b,p=ab$, the equations become $$\tag1s^2-2p+s=12$$ $$\tag2p+s=3$$ To get rid of $p$, combine $(1)+2(2)$: $$s^2+3s=18$$ So either $s=3$ (and $p=0$) or $s=-6$ (and $p=9$). At any rate, $a,b$ are the roots of $x^2-sx+p$, so the solutions are $$(a,b)=(0,3) \text{or} (3,0) \text{or} (-3,-3).$$