a=kb+m. 3^a=(-2)^k*3^m (mod 3^b+2). So 3^b+2 divides 2^(k-1)*3^m+(-1)^k which divides 2^(k-1)*3^b+(-1)^k*3^(b-m). this is equivalent to 3^b+2 divides 2^k-(-1)^k*3(b-m). this is clearly not 0, so it is at least 3^b+2 (since it is at least 2^k-3^b that is bigger than -(3^b+2) since m>=0). from triangle inequality 2^k+3^(b-m)>=3^b+2. If m=0, then either k=1 so a=b contradiction, or k>1, so 2^k>(3^b) so k>b so a>b^2. If m>=1, 2^k>=2*(3^(b-1))+2, which for b>=2 implies k>=b, so a>=b^2+1 and we are done. b=1 is trivial.

What ?!!!! Obviously $a>b$ Let $a=bx+y$ where $0\le y\le b-1$ Claim: $x\ge b$ Suppose by contradiction that $x\le b-1$ Then $3^b+2\mid (3^b)^x\cdot 3^y+2$ so $3^b+2\mid (-2)^x\cdot 3^y+2$ so $3^b+2\mid (-2)^x\cdot 3^y-3^b$.Since $b>y$ and $(3,3^b+2)=1$ we have $3^b+2\mid (-2)^x-3^{b-y}$.Divide into cases:(and also note $(-2)^x\neq 3^{b-y}$) Case 1: $x$ even In this case we have $3^b+2\mid 2^x-3^{b-y}$. If $2^x>3^{b-y}$ then $3^b>2^b>2^{b-1}\ge 2^x\ge 3^{b-y}+3^b+2>3^b$ contradiction. If $3^{b-y}>2^x$ then $3^b\ge 3^{b-y}\ge 2^x+3^b+2>3^b$ another contradiction Case 2: $x$ odd then $2^x+3^{b-y}\ge 3^b+2$.If $y>0$ then $2^{b-1}+3^{b-1}\ge 2^x+3^{b-y}\ge 3^b+2$ so $2^{b-1}>2\cdot 3^{b-1}>3^{b-1}\ge 2^{b-1}$ contradiction. Thus we must have $y=0$ i.e $3^b+2\mid 2^x+3^b$ or $3^b+2\mid 2^x-2$.If $x=1$ we have $a=b$; is $x>1$ then $3^b<3^b+2\le 2^x-2<2^x<2^b<3^b$ false $\blacksquare$ By the claim we have $a\ge b^2$ so we must have $a=b^2 $ meaning $3^b+2\mid (-2)^b-3^b$ or $3^b+2\mid (-2)^b+2 $ which by the same boundings as above it fails. We are done !

Let's suppose, FTSOC, that $a\leq b^{2}$. Let $a=mb+k$, where $k$ is the remainder when $a$ is divided by $b$. Suppose that $m \leq b$. We have $3^{b}+2|3^{mb+k}+2$, and also $3^{b}+2|3^{mb+k}+2.3^{(m-1)b+k}$. Subtracting these two, we get $3^{b}+2|3^{(m-1)b+k}-1$(we can divide it by $2$, because $3^{b}+2$ is odd. We can continue this process and we get: $3^{b}+2|2.3^{(m-2)b+k}+1$ $3^{b}+2|4.3^{(m-3)b+k}-1$ $3^{b}+2|8.3^{(m-4)b+k}+1$ Well, observing this we can guess that $3^{b}+2|2^{l-1}.3^{(m-l)b+k}+(-1)^{l}$ for $l=1, 2, ..., m$. This easily follows by induction.

Now, when we proved this, take $l=m$. We have $3^{b}+2|2^{m-1}.3^{k}+(-1)^{m}$ Since $a$ and $b$ are distinct, at least one of the following two is not true: $m=1$ and $k=0$, hence $2^{m-1}.3^{k}+(-1)^{m}$ cannot be $0$. Also, $m$ cannot be $0$, because obviously $a>b$. We have 2 cases now: Case 1: $m$ is even. Then we have $3^{b}+2|2^{m-1}.3^{k}+1$ Clearly, $b\geq k$, so we can multiply this by $3^{b-k}$ and we get $3^{b+2}|2^{m-1}.3^{b}+3^{b-k}$. We also have $3^{b}+2|2^{m-1}.3^{b}+2^{m}$ . Subtracting these two, we get $3^{b}+2|3^{b-k}-2^{m}$. This is clearly not $0$, so we have two cases: Case 1.1:$3^{b-k}-2^{m}>0$. But obviously $3^b+2>3^{b-k}-2^{m}$, which is a contradiction. Case 1.2:$3^{b-k}-2^{m}<0$. We have $2^{m}-3^{b-k} \leq 2^{b}-3^{b-k}<3^{b}-3^{b-k}<3^{b}+2$, contradiction! Case 2: $m$ is odd. Then we have $3^{b}+2|2^{m-1}.3^{k}-1$. Similarly to Case 1, we get $3^b+2|3^{b-k}+2^{m}$. Well, this is obviously positive. If $k=0$, we get $3^{b}+2|2^{m-1}-1$. $3^{b}+2>2^{m-1}-1$, so the only option is $2^{m-1}=1$, which implies $m=1$ and $a=b$. So $k \geq 1$. We will prove that $3^{b}+2>3^{b-k}+2^{m} \iff 3^{b}-3^{b-k}>2^{m}-2$, which will be a contradiction. LHS $\geq 3^{b}-3^{b-1}=2.3^{b-1}>2.(2^{b-1}-1)\geq 2.(2^{m-1}-1)=$RHS. Hence we have contradiction with the assumption that $m\leq b$. Hence, $m>b$, and then $a>b^{2}$, as desired. $\blacksquare$ Remark: I saw the problems thirty minutes before they were posted, and immediately started solving this one(because I found it interesting), and I wanted to be the first poster . When I saw they had been posted, I started writing a solution, and, as I don't have much experience with $\LaTeX$, it took me very long to write this post, and in the process, two people posted their solutions(which are similar to mine), but I decided to post it anyway.

Way too easy for P3. Here is an alternative with a simpler ending. Let $a=bk+r$ with $0\le r\le b-1$. Then $3^a+2 = 3^{kb+r}+2\equiv (-2)^k 3^r+2\pmod{3^b+2}$. Hence, $3^b+2\mid (-2)^k3^r +2 - 3^b-2 = 3^r\left((-2)^k -3^{b-r}\right)$. Since $3^b+2$ and $3^r$ are coprime, we thus arrive at \[ 3^b+2\mid (-2)^k -3^{b-r}. \]Notice that since $b-r\ge 1$, $(-2)^k-3^{b-r}$ is never zero. Suppose first $k$ is even. Then, $3^b+2\mid 2^k-3^{b-r}$. This immediately yields $k>b$ so $a=kb+r>b^2$. Now let $k$ be even. If $r=0$ then $3^b+2\mid 2^k-2$, yielding once again $k>b$, so suppose $r\ge 1$. Notice that \[ 3^b+2\mid 2^k+3^{b-r}\Rightarrow 2^k\ge 3^{b-r}\left(3^r-1\right)+2\ge \frac{2}{3} 3^b +2 \Leftrightarrow 2^{k-1}\ge 3^{b-1}+1 \]using $3^r-1\ge 2\cdot 3^{r-1}$ valid for $r\ge 1$. But this precisely yield $k>b$, so $a>b^2$.

This problem was proposed by me

Let $a=xb+y$ with $x$ and $y$ being non negative numbers with $x$ larger than $0$ and $y$ smaller than $b$. Polynomial division gives that $3^{b}+2|3^{xb}-(-2)^x|3^{xb+y}-(-2)^x3^y$. So we must have $3^b+2|2+(-2)^x3^y\Rightarrow$ $3^b+2|(-2)^x3^y-3^b\Rightarrow 3^b+2|(-2)^x+3^{b-y}$. If $y=0$ then we must have $3^b+2|(-2)^x-2$ so we must have $b<x$. If $y\neq 0$ then $(-2)^x+3^{b-y}\leq 3^{b-1}+3^{x}$ so we must have at least $b\leq x$. This concludes the proof. Remark: This solution was motivated by the fact that the polynomial $x^n+2$ is irreducible and that the inequality is by no means sharp.

Tynyshbek agay (teacher) finally getting his recognition for his math competence, years of working and beautiful method of teaching. He was an algebra teacher at preparation camps for Olympiads. And I am happy that I was able to be his student:)))

Very instructive problem for people of all ages! Since $a\neq b$ and $3^b + 2 \leq 3^a + 2$, we must have $k = a - b > 0$. We have $3^b + 2 \mid 3^a - 3^b$ and thus $3^b + 2 \mid 3^k - 1$. Working further on the aim to reduce the numerator as much as possible, write $k = bq+r$ where $0 \leq r < k$. Then $1 \equiv 3^k \equiv 3^{bq+r} \equiv (-2)^q \cdot 3^r \pmod {3^b+2}$, i.e. $3^b + 2 \mid 2^q \cdot 3^r \pm 1$. One might stop here and get stuck (I did so for a few minutes when solving this), but notice that we can make the numerator even smaller as follows: $3^b +2 \mid 2^q \cdot 3^b \pm 3^{b-r}$, i.e. $3^b +2 \mid 2^{q+1} \pm 3^{b-r}$. Note that the latter numerator is non-zero e.g. by parity reasons. If $r>0$, then from the latter divisibility we have $3^b < 3^b +2 \leq \left|2^{q+1} \pm 3^{b-r}\right| = 2^{q+1} + 3^{b-r} \leq 2^{q+1} + 3^{b-1}$, thus $2 \cdot 3^{b-1} \leq 2^{q+1}$, so $q > b - 1$, implying $a = b + k = b(q+1) + r > b(q+1) > b^2$. If $r=0$, then $3^b + 2 \mid 2^{q+1} \pm 3^b$, so $3^b + 2 \mid 2^{q+1} \pm 2$, i.e. $3^b + 2 \mid 2^{q} \pm 1$ (with the numerator being non-zero), thus again $3^b < 3^b + 2 \leq 2^{q} + 1 \leq 3^q$, i.e. $q>b$, implying $a = b + k = b(q+1) + r > bq > b^2$.

Problem finishes with $a=bx+y$

We uploaded our solution https://calimath.org/pdf/BMO2024-3.pdf on youtube https://youtu.be/5-GG2ai1FNc.

Let $a=:bq+r$ for nonnegative integers $q$ and $r<b$. We can prove by induction that $3^b+2\mid 2^{q-1}3^r+(-1)^q$. If $r=0$ then $2^{q-1}>3^b$ so $q>b+1$ and we are done. So assume $r>1$. Let $n:=\frac{2^{q-1}3^r+(-1)^q}{3^b+2}$ so $2n\equiv (-1)^q\pmod{3^r}$. It follows that $n\equiv\frac{3^r+(-1)^q}{2}\pmod{3^r}$ so $n\geq\frac{3^r-1}{2}$. Now \[ (3^b+2)(3^r-1)\leq 2^q3^r+2 \]so \[ 2^q\geq 3^b-3^{b-r}+2-\frac{4}{3^{r-1}}\geq 3^{b-1}. \]Thus $q\geq b$, as desired. $\square$

https://drive.google.com/file/d/1Tbb2CdkE4b2Gr3w32l68RJB4hsS1MiII/view?usp=drivesdk this is my solution to this problem click on it and see my solution.https://drive.google.com/file/d/1Tbb2CdkE4b2Gr3w32l68RJB4hsS1MiII/view?usp=drivesdk.you have to do that thing when u coopy my solution.then u can open link.

Clearly, $a>b$ so let $a=bk+r$, where $k\ge 1$ and $0\le r\le a-1$ and suppose that $k\le b-1$. We have that $3^b+2\mid 3^{bk+r}+2$ but notice that $3^{bk+r}=\left(3^b\right)^k\cdot 3^r\equiv(-2)^k\cdot3^r\pmod{3^b+2}$. Therefore we get $$3^b+2\mid(-2)^k\cdot3^r+2\iff 3^b+2\mid (-2)^k-3^{b-r}$$ Now, if $r\neq 0$ we get $3^b+2\le |(-2)^k-3^{b-r}|\le 2^k+3^{b-r}\le 2^{b-1}+3^{b-1}$ which is clearly false. Otherwise, if $r=0$ we get $3^b+2\mid (-2)^k+2$ so $3^b+2\le |(-2)^k+2|\le 2^k+2\le 2^{b-1}+2$ which is also false. $\blacksquare$

Nice Algebra question!. Suppose FTSOC that $a=kb+\ell \le b^2$ for some $k \ge 1$ and $b>\ell \ge 0$, (where if $k=1$ then $\ell \ge 1$), then note that $3^b+2 \mid 3^a-3^b=3^b(3^{a-b}-1)$ and as $\gcd(2,3)=1$ we have that $3^b+2 \mid 3^{a-b}-1=3^{(k-1)b+\ell}-1$ so $3^b+2 \mid (-2)^{k-1} \cdot 3^{\ell}-1$ and thus by multiply by $(-1)^{k-1}$ we know that $3^b+2 \mid 2^{k-1} \cdot 3^{\ell}+(-1)^k$ now multiplying by $3^{b-\ell}$ gives $3^b+2 \mid 2^k+(-1)^{k-1} \cdot 3^{b-\ell}$. Now in the last one RHS can't be zero so it's at least $3^b+2$, also from the assumption we have that $b>k$, now if $k$ was odd then we have that $3^b+2 \le 2^k+3^{b-\ell}$, if $\ell \ge 1$ then this gives $2^k \ge 3^{b-1}+2 \ge 3^k+2$, contradiction!, so actually $\ell=0$ has to be true in this case, but then we had $3^b+2 \mid 2^{k-1}-1$ which clearly implies $k \ge b+1$, contradiction!. If $k$ is even then we have two cases which we will delete now. Case 1: $3^{b-\ell}>2^k$ In this case we get $3^{b-\ell} \ge 3^b+2+2^k$, clearly a contradiction. Case 2: $2^k>3^{b-\ell}$. Then in fact we have that $3^k>2^k \ge 3^{b-\ell}+3^b+2>3^b$ which means $k>b$ again, a contradiction!. Thus in all thecases we get a contradiction, thus we are done .