Problem 8. For what values of the parameter $a$ does the equation

$$
3^{x^{2}+6 a x+9 a^{2}}=a x^{2}+6 a^{2} x+9 a^{3}+a^{2}-4 a+4
$$

have exactly one solution?
Answer: Only when $a=1$.

Solution. Let's denote $x+3a$ by $t$. Note that the number of solutions to the equation does not change with this substitution. Then the original equation will take the form

$$
3^{t^{2}}=a t^{2}+a^{2}-4 a+4
$$

Notice that the expressions on both sides do not change when $t$ is replaced by $-t$, so the equation can have an odd number of solutions (in particular, exactly one solution) only if $t=0$ is a root:

$$
3^{0}=a \cdot 0+a^{2}-4 a+4
$$

i.e., $a^{2}-4 a+3=0$, from which $a=3$ or $a=1$. Thus, besides these two numbers, no other values of the parameter $a$ can satisfy the condition.

Let $a=1$. Then the equation will take the form $3^{t^{2}}=t^{2}+1$. Note that $3^{x}>x \ln 3+1$ for $x>0$ (which can be proven, for example, by taking the derivatives of both sides). Then for $t \neq 0$ we get $3^{t^{2}}>t^{2} \ln 3+1>t^{2}+1$. Therefore, when $a=1$, the equation has a unique solution.

Let $a=3$. Then the equation will take the form $3^{t^{2}}=3 t^{2}+1$. Note that $3^{1}=3, 3 \cdot 1+1=4$, but $3^{2^{2}}=81$, $3 \cdot 2^{2}+1=13$, i.e., at $t=1$ the left side is less than the right, and at $t=2$ the opposite is true. Therefore, by the intermediate value theorem, the equation has at least one more root in the interval $(1,2)$. Therefore, $a=3$ does not satisfy the condition.