41st Balkan Mathematical Olympiad

Obviously we have $a>b$. Let $a=bq+r$, where $0\le r<b$. Then $$3^a\equiv 3^{bq+r}\equiv (-2{)}^q\cdot 3^r\equiv -2(\mathrm{mod}{3}^b+2)$$ So $3^b+2$ divides $A=(-2{)}^q\cdot 3^r+2$ and it follows that $$|(-2{)}^q\cdot 3^r+2|\ge 3^b+2\text{ or }(-2{)}^q\cdot 3^r+2=0.$$ We make case distinction:

1. $(-2{)}^q\cdot 3^r+2=0$. Then $q=1$ and $r=0$ or $a=b$, a contradiction.

2. $q$ is even. Then $$A=2^q\cdot 3^r+2=(3^b+2)\cdot k.$$ Consider both sides of the last equation modulo $3^r$. Since $b>r$: $$2\equiv 2^q\cdot 3^r+2=(3^b+2)k\equiv 2k(\mathrm{mod}{3}^r),$$ so it follows that $3^r|k-1$. If $k=1$ then $2^q\cdot 3^r=3^b$, a contradiction. So $k\ge 3^r+1$, and therefore: $$A=2^q\cdot 3^r+2=(3^b+2)k\ge (3^b+2)(3^r+1)>3^b\cdot 3^r+2$$ It follows that $$2^q\cdot 3^r>3^b\cdot 3^r,\text{ i.e. }{2}^q>3^b,\text{ which implies }{3}^{b^2}<2^{bq}<3^{bq}\le 3^{bq+r}=3^a.$$ Consequently $a>b^2$.

3. If $q$ is odd. Then $$2^q\cdot 3^r-2=(3^b+2)k.$$ Considering both sides of the last equation modulo $3^r$, and since $b>r$, we get: $k+1$ is divisible by $3^r$ and therefore $k\ge 3^r-1$. Thus $r>0$ because $k>0$, and: $$2^q\cdot 3^r-2=(3^b+2)k\ge (3^b+2)(3^r-1),\text{ and therefore}$$ $$2^q\cdot 3^r>(3^b+2)(3^r-1)>3^b(3^r-1)>3^b\frac{3^r}{2},\text{ which shows}$$ $$2^q+1>3^b.$$ But for $q>1$ we have $2^q+1<3^q$, which combined with the above inequality, implies that $3^{b^2}<(2^q+1{)}^b<3^{qb}\le 3^a$, q.e.d. Finally, If $q=1$ then $2^q\cdot 3^r-2=(3^b+2)k$ and consequently $2\cdot 3^r-2\ge 3^b+2\ge 3^{r+1}+2>2\cdot 3^r-2$, a contradiction.

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Alternative solution.

$D=a-b$, and we shall show $D>b^2-b$. We have $3^b+2|3^a+2$, so $3^b+2|3^D-1$. Let $D=bq+r$ where $r<b$. First suppose that $r\ne 0$. We have $$1\equiv 3^D\equiv 3^{bq+r}\equiv (-2{)}^{q+1}{3}^{r-b}(\mathrm{mod}{3}^b+2)⟹3^{b-r}\equiv (-2{)}^{q+1}(\mathrm{mod}{3}^b+2)$$ Therefore $$3^b+2\le |(-2{)}^{q+1}-3^{b-r}|\le 2^{q+1}+3^{b-r}\le 2^{q+1}+3^{b-1}$$ Hence $$2\times 3^{b-1}+2\le 2^{q+1}⟹3^{b-1}<2^q⟹\frac{\log 3}{\log 2}(b-1)<q$$ Which yields $D=bq+r>bq>\frac{\log 3}{\log 2}b(b-1)\ge b^2-b$ as desired. Now for the case $r=0$, $(-2{)}^q\equiv 1(\mathrm{mod}{3}^b+2)$ and so $$3^b+2\le |(-2{)}^q-1|\le 2^q+1⟹3^{b-1}<3^b<2^q⟹\frac{\log 3}{\log 2}(b-1)<q$$ and analogous to the previous case, $D=bq+r=bq>\frac{\log 3}{\log 2}b(b-1)\ge b^2-b$. □