I think I got this... hopefully... We find the solutions (1, 1), (27, 3), and (16, 2), and we claim that no others exist. First of all, a and b must have the same ratios of prime factors, so one must divide the other. Note also that if b = xa, x > 1, then $a^{x^2a^2} = (xa)^a \implies a^{x^2a} = xa \implies a^{x^2a - 1} = x$ Since x > 1, $a \ge 2$. If x = 2, then $a^{2^2a-1} \ge 2^{2 \cdot 2^2-1} = 2^7 > 2$. Then we can show easily by induction that if x > 2, $a^{x^2a-1} > x$. We now deal with the case $a \ge b$. If $a = b$, $a^{a^2} = a^a \implies a = 1$ gives us the solution (1, 1). Let $a = kb, k > 1$. $(kb)^{b^2} = b^{kb}$ $k^{b^2} = b^{kb-b^2}$ $k^b = b^{k-b}$ Note that $k-b > 0 \implies k > b \implies b | k$. Let $k = mb$. $(mb)^b = b^{(m-1)b}$ $mb = b^{m-1}$ $m = b^{m-2}$ Now test cases m = 1, 2, 3, 4. It turns out that m = 3, b = 3 works, giving us the solution (27, 3), and m = 4, b = 2 works, giving us the solution (16, 2). If m = 5, since $b \ge 2$, we have $b^{5-2} \ge 8 > 5$. From there on, it is easy to see by induction that $b^{m-2} > m$, so no other solutions exist.

My solution seems to be the same with some differences where I have no much number theory in it .. First we note that $(1,1)$ is a solution , so we can assume that $a \neq 1 $ , so we take the logarithms of base $a$ to have \[ b^2 = a \log_a b \] , but then clearly $\log_a b $ is a rational number. Hence we can write it in lowest terms $\log_a b = \frac{m}{n} $ which imply $b = a^{\frac{m}{n}}$ , so that we have \[ a^{2m/n} = \frac{am}{n} \] , We prove that $m=1$ , if not we treat the two cases $n\geq 2m $ and $ n < 2m $. If $n \geq 2m$ ,we raise the equation to the power $n$ and then divide by $a^n$ to have \[ 1 = a^{n-2m} \frac{ m^n}{n^n} \] , but this is absurd as then $m^n$ can't cancel with any factor of the denominator.If $n < 2m $ ,one can obtain \[ a^{2m-n}=\frac{m^n}{n^n} \] , but then $n=1$ , and then $a^{2m-1}=m$ which is a contradiction. So we have $\log_a b = 1/n $ , and then $b^n = a $ , substituting in the original equation we have \[ b^{nb^2} = b^{b^n} \]. We take logarithms again to have \[ n = b^{n-2} \] , which obviously have solutions $(b,n)=(2,4),(3,4)$ only , and then original equation has $(a,b) \in \{(1,1),(2^4,2),(3^3,3)\}$

Take gcd(a,b)=d so that a=ud, b=vd and gcd(u,v)=1 ; and then use induction on variables to limit the possibilities of the values that the variables can take (a widely known technique in Number Theory)

@Nanas Why m is equal to 1

@oiviluhatef can you express the technique in detail??

If I got to the equation $b^{k-1} = k+1,$ could I just say "it is obvious that the only solutions for $(b,k)$ are $(1,0),(2,3),(3,2),$ or would I have be rigorous by taking derivatives?

I would give at least an explanation on an olympiad (but probably take derivatives) to make sure I don't lose points.

So basically,this problem must be divided into 3 cases: $ a \geqslant b > 1$ ,$b \geqslant a > 1$ and one of them is 1(the no-brainer one). First we shall do $ a \geqslant b$: $ a^{b^2} = b^a $,from this we see that $a = b^t$,inserting that back into our beginning equation we have: $$ b^{tb^2}= b^{b^t} $$$$ tb^2 = b^t $$$$ t = b^{t-2} $$When $ b \geqslant 4 $ we have the following $ b^{t-2} \geqslant 4^{t-2} > t $ This can be easily seen by induction: Our base case $t=3$ works,thus let's say that for some $t$ the inequality holds then we have the following: $4^{t-2} > t \implies 4^{t-1} > 4t > t+1 $, thus the inequality holds by induction.Thus we have that $1 < b < 4$. When $b=2$ we have $t = 2^{t-2}$,we see that $t=4$ (also can be seen by inequalities,which can be shown by induction). When $b=3$ we have that $t=3^{t-2}$,we see that $t=3$ ( also seen by inequalities,which can be shown by induction). Now let's tackle the case $b \geqslant a $: Again we see that $b = a^t$,thus we have the following: $$ a^{a^{2t}} = a^{at} $$Thus we have that $a^{2t} = at$,this is simply not possible for $a > 1$.because by induction: $$ a^{2t-1} > t $$$$a^{2t} > at > t+1 $$Of course the base case $a = 2$ works Now the no-brainer $a =1 $ Now we have that $b^1 = 1$,thus $b=1$ So the solutions are: $$\boxed{(a,b) \in \{(1,1);(16,2);(27,3) \}}$$

Does this equation have negative roots?

Birdews wrote: Does this equation have negative roots? What about $a=b=-1$?

Solution from Twitch Solves ISL: The answer is $(1,1)$, $(16,2)$ and $(27,3)$. We assume $a,b > 1$ for convenience. Let $T$ denote the set of non perfect powers other than $1$. Claim: Every integer greater than $1$ is uniquely of the form $t^n$ for some $t \in T$, $n \in {\mathbb N}$. Proof. Clear. $\blacksquare$ Let $a = s^m$, $b = t^n$. \[ s^{m \cdot (t^n)^2} = t^{n \cdot s^m}. \]Hence $s = t$ and we have \[ m \cdot t^{2n} = n \cdot t^m \implies t^{2n-m} = \frac nm. \]Let $n = t^e m$ and $2 \cdot t^e m - m = e$, or \[ e + m = 2t^e \cdot m. \]We resolve this equation by casework If $e > 0$, then $2t^e \cdot m > 2e \cdot m > e+m$. If $e=0$ we have $m=n$ and $m = 2m$, contradiction. If $e = -1$ we apparently have \[ \frac{2}{t} \cdot m = m-1 \implies m = \frac{t}{t-2} \]so $(t,m) = (3,3)$ or $(t,m) = (4,2)$. If $e = -2$ we apparently have \[ \frac{2}{t^2} \cdot m = m - 2 \implies m = \frac{2}{1 - 2/t^2} = \frac{2t^2}{t^2-2}. \]This gives $(t,m) = (2,2)$. If $e \le -3$ then let $k = -e \ge 3$, so the equation is \[ m-k = \frac{2m}{t^k} \iff m = \frac{k \cdot t^k}{t^k-2} = k + \frac{2k}{t^k-2}. \]However, for $k \ge 3$ and $t \ge 2$, we always have $2k \le t^k - 2$, with equality only when $(t,k) = (2,3)$; this means $m=4$, which is not a new solution.

The answer is $(27, 3), (16, 2), (1, 1)$. It is easy to see that if $a^m=b^n$ and $a>b$ then $a=b^k$. $\textbf{Case 1:}$ $a > b$ Then $a = b^k$. Substituting we get \[ b^{k^{b^2}}=b^{b^2k}=b^{b^k} \implies b^2k=b^k \implies k=b^{k-2} \]Notice that $1^{k-2} < k < 2^{k-2}$ if $k > 4$ which means that we only need to test out $k$ from $[2, 4]$. Testing out each each $k$ we find the only solutions are $(27, 3), (16, 2)$. $\textbf{Case 2:}$ $a=b$ This is trivial, so I won't go into detail. In the end we get $(1, 1)$. $\textbf{Case 3:}$ $a < b$ This means that $b = a^k$. Doing the same steps as above we get $k=a^{2k-1}$. Notice that $1^{2k - 1} < k < 2^{2k - 1}$ if $k > 1$. Now if $k=1$ then we have Case $2$ which we already covered. Thus we can conlude that these are the only solutions.

The solutions are $(1,1)$, $(16,2)$, and $(27,3)$. We have $a=1$ if and only if $b=1$. Notice that $a=2$ has no solutions. If $b=2$, then $a$ is a power of $2$. Quick casework gives $a=16$. From now on, assume that $a,b \geq 3$. Then, the equation implies that $a^b<b^a$. It's well known (by calculus) that this implies $a>b$ when $a,b \geq 3$. Notice that $a^\frac{b^2}{a}=b$, so there must be nonnegative integers $n$, $c$, and $d$ such that $a=n^c$ and $b=n^d$. Notice that $c>d$. Then, we have $cn^{2d}=dn^c$ or $n^{c-2d}=\frac{c}{d}$. Now, we do casework on $c-2d$. Since $c>d$, we know that $c-2d$ is positive. Then, it's pretty easy to see by bounding that the only solution when $a,b \geq 3$ is $(27,3)$.

Beautiful

When in doubt, calcbash an inequality to prove that a finite number of solutions exist. The answer is $(a,b)=(1,1),(16,2),(27,3)$ only, which can be verified to work, so we only have to show nothing else works. Hence suppose $a,b>1$. Clearly for any prime $p$, we have $p \mid a \iff p \mid b$. Further, we can verify that $a=b^2$ doesn't work, as if $a=b^2$ then $$a^a=b^a \implies a=b=b^2 \implies b=1,$$which is impossible. Hence we can consider the following two cases: Case 1: $a>b^2$. Then, we have $$b^a=a^{b^2}>b^{2b^2} \implies a>2b^2.$$Next let $p$ be a prime dividing $a$ and $b$, so we have $$b^2\nu_p(a)=a\nu_p(b) \implies \frac{\nu_p(a)}{\nu_p(b)}=\frac{a}{b^2}>2,$$so we have $\nu_p(a)>2\nu_p(b) \implies b^2 \mid a$. Thus letting $\tfrac{a}{b^2}=k$ where $k>2$ is a positive integer, we have $\nu_p(a)=k\nu_p(b) \implies a=b^k$. Substituting this, we obtain $$b^{kb^2}=b^{b^k} \implies b^{k-2}=k.$$I claim that for $b \geq 4$, we actually have $b^{k-2}>k$ for all $k\geq 3$. This is true for $k=3$, and since we have $b\geq 4>2>\tfrac{k+1}{k}$, by induction the inequality holds for all $k$. Thus we only have to check $b=2,3$. For $b=2$ we get $2^k=4k$, which gives $k=4$ as the only solution—no solutions exist for $k<4$, and for $k>4$ we can use the fact that $2>\tfrac{k+1}{k}$ to prove that $2^k>4k$. Similarly, for $b=3$ we get $3^k=9k$ which gives $k=3$ as a solution, nothing for $k<3$, and $3^k>9k$ for $k>4$. These result in the solutions $(16,2),(27,3)$. Case 2: $a<b^2$. Then, we have $$b^a=a^{b^2}>a^a \implies a<b.$$Further, we have $b^a=a^{b^2}>a^b \implies \sqrt[b]{b}>\sqrt[a]{a}$. But we can verify that $f(x)=x^{1/x}$ is decreasing past $x>e$, since $$\frac{d}{dx}\ln(f(x))=\frac{d}{dx}\frac{\ln(x)}{x}=\frac{1-\ln(x)}{x^2}<0$$for $x>e$ by the quotient rule, so $\ln(f(x))$ is decreasing past that point and so is $f(x)$. Thus if $b>a\geq 3$ and $a<b$, we have $\sqrt[a]{a}>\sqrt[b]{b}$. Therefore we require $a=2$, in which case the original equation becomes $2^{b^2}=b^2$. On the other hand, for all positive integers $k$ we have $2^k>k$, since the inequality is true for $k=1$ and $2\geq \tfrac{k+1}{k}$ so the inequality holds for all $k$, so this case yields no solutions. Combining these two cases implies that no solutions other than the three given exist, as desired. $\blacksquare$

The solutions are $(a,b)=(1,1),(27,3),(16,2)$. It's obvious that $b$ is a rational power of $a$. Set $b=a^r$; we can rearrange to get $a^{r^2a^2}=r^aa^a$, or $a^{r^2a-1}=r$. Here, it's clear again that $r$ is a rational power of $a$. If $a=1$, it's immediate that $b=1$ which gives the first pair of solutions. Now, assume $a\ge 2$; let $r=a^k$, and take the log base $a$ of both sides gives $a^{1+2k}=k+1$. Here, we consider 3 cases: If $k\ge 0$, we have $\frac{d}{dk} a^{1+2k}=2\cdot a^{1+2k}\ln(a)>1=\frac{d}{dk} (k+1)$ and $a^1>1$ when $k=0$, so there are no solutions in this case. If $k\le -1$, a real power of a positive number would be nonpositive, which is impossible. Now, suppose that $k\in (-1,0)$. Since $1+2k$ and $k+1$ are rational, the LHS and RHS necessarily have the form $\frac{1}{n}$ where $c=n^{-2k-1}$ is a positive integer and a rational power of $a$. Let $k=\frac{1-n}{n}$; we can derive $a^{\frac{2-n}{n}}=\frac{1}{n}$, and hence $n$ is a $|2-n|=n-2$th power. Via inspection, this has solutions at $n=3,4$ which correspond to the last two ordered pairs $(a,b)$. Claim: $2^n>4n$ for $n\ge 5$. Proof: We proceed by induction. The base case $n=5$ is straightforward. For the inductive step, assume that $2^{n-1}>4n-4$; after multiplying both sides by $2$, we get $2^n>8n-8$, and since $n\ge 2$ it holds that $8n-8\ge 4n$. Therefore, the numeric solutions we have found are exhaustive and the proof is complete.

It is easy to see that $b=1\implies a=1$ $b=2 \implies a=16$ and $b=3\implies a=27$ (by considering the exponential functions and proving monotonicity by taking derivatives). We will prove that those are the only solutions. Assume $b>3$. It is not hard to check that $a\neq 1$ and $a\neq 2$. So, we have $a\geq 3$. If $a\leq b$, since $e<a\leq b$, $a^b\geq b^a=a^{b^2}>a^b$, which is impossible. So, we have $a>b$. Then, there must exist prime $p$ such that $v_p(a)>v_p(b)$. Takin vp of our equation, $b^2v_p(a)=av_p(b)$ $\implies a>b^2$. But for any prime $q$, $b^2v_q(a)=av_q(b)$ $\implies v_q(b)<v_q(a)$ for any prime $q$, equivalently $b\mid a$. Let $a=kb, k>b$. The equation becomes $k^b=b^{k-b}$. Since $k>b$, $b<k-b$ or $k>2b$. Then, $a>2b^2$ and for any prime $p$, $\frac{v_p(a)}{v_p(b)}=\frac{a}{b^2}>2$. This gives $b^2\mid a$ (as $2v_p(b)<v_p(a)$). Let $d=\frac{a}{b^2}$. We have $\frac{v_p(a)}{v_p(b)}=d$ for any prime $p$ or $v_p(a)=v_p(b^d)$ for any prime $p$, which is equavalent to $a=b^d$. So, substituting that into our equation, $(b^d)^{b^2}=b^{b^d}$. This reduces to $d=b^{d-2}$. Clearly, $d\geq 3$. Finally $d=(1+(b-1))^{d-2}\geq 1+(b-1)(d-2)$ by Bernoulli's inequality. This gives $d\geq 1+(b-1)(d-2)$, which is equivalent to $$1\geq (d-2)(b-2)$$This is a contradiction since $b-2>1$ and $d-2\geq 1$

The answer is $(a,b) = (1,1),(16,2),(27,3)$. Observe $a,b$ must have the same set of primes divisors. Assume $a,b \ne 1$. Write $$a = p_1^{\alpha_1} \cdots p_k^{\alpha_k} \qquad , \qquad b = p_1^{\beta_1} \cdots p_k^{\beta_k} $$Then the given condition is equivalent to $$ t = \frac{\alpha_1}{\beta_1} = \cdots = \frac{\alpha_k}{\beta_k} = \frac{a}{b^2} = p_1^{\alpha_1 - 2\beta_1} \cdots p_k^{\alpha_k - 2 \beta_k} $$Note all of $\alpha_1 - 2 \beta_1, \ldots,\alpha^k - 2 \beta_k$ have the same sign. In particular either $\alpha_i \mid \beta_i$ for all $i$ OR $\beta_1 \mid \alpha_i$ for all $i$. Claim: $\alpha_1 > \beta_1$. Proof: FTSOC $\beta_1 \ge \alpha_1$. Then $$ \beta_1 \ge \frac{\beta_1}{\alpha_1} = p_1^{2 \beta_1 - \alpha_1} \cdots p_k^{2 \beta_k - \alpha_k} \ge p_1^{2 \beta_1 - \alpha_1} \ge p_1^{\beta_1} \ge 2^{\beta_1} \ge \beta+1 $$Thus we obtain a contradiction. $\square$ Now $t$ is an integer, and moreover $t \ge 2$. We in fact have $t \ge 3$. Claim: $t \le 4$. Proof: Observe $$t = \frac{\alpha_1}{\beta_1} \ge p_1^{\alpha_1 - 2 \beta_1} \ge 2^{t - 2} $$Thus $t \le 4$. $\square$ Now it is easy to check that for $t=3,4$, the only solutions are $(a,b) = (27,3),(16,2)$, respectively. $\blacksquare$

ilikemath40 wrote: we only need to test out k from [2, 4]. Testing out each each k we find the only solutions are (27, 3), (16, 2). I don't understand this part : How do you prove that the only solutions for k in ]1;4] are in [2;4] and that the solutions are then only the two cases (27,3) and (16,2) ?

Anyone ?

First, writting it as a logarithm we get $b^2 = a \log_a b$ Because $(a,b)$ are positive integers, we get that $\log_a b$ is a rational number Let $\log_a b = \frac{c}{d}$, so $b = a^{\frac{c}{d}}$. We get $a^{a^{\frac{2c}{d}}} = a^{\frac{ac}{d}}$ which implies $a^{\frac{2c}{d}} = \frac{ac}{d}$ From this, we can prove that $c=1$ and $b^d = a$ Plug it back in to get $b^{db^2} = b^{b^d}$, which implies $d=b^{d-2}$ We then have solutions of $(1,1)$, $(16,2)$, and $(27,3)$

The only solutions are $(1,1)$, $(16,2)$, and $(27,3)$, which all clearly work. Now we prove they are the only solutions. Notice that \[\frac{\nu_p(b)}{\nu_p(a)} = \frac{b^2}{a},\]so either $a\mid b$ or $b\mid a$. If $a=b$, then $a = a^2$, so $a=b=1$, so now assume $a\ne b$, and also that $a,b>1$. Case 1: $a\mid b$. Let $b = k\cdot a$, where $k>1$. We have $a^{k^2a^2} = k^a\cdot a^a$, so $a^{k^2 a - 1} = k$. However, we have \[a^{k^2a - 1} > a^{2k^2-1} \ge 2^{k^2} >k,\]contradiction. Case 2: $b\mid a$. \ Let $a = k\cdot b$, where $k>1$. We have $b^{b^2} k^{b^2} = b^{kb},$ so $b^{k-b} = k^b$. This implies that $k>b$ because $b^{k-b} = k^b>1$. Now, $k-b>b$ because otherwise $b^{k-b} < b^b < k^b$. Notice that for any prime $p$, we have $\frac{\nu_p(k)}{\nu_p(b)} = \frac{k-b}{b}>1$, so $b\mid k$. Let $k = mb$, where $m>1$. We get \[b^{b(m-1)} = (mb)^b \implies b^{m-2} = m\]If $m>4$, then $b^{m-2} \ge 2^{m-2} > m$, contradiction, so $m\in \{2,3,4\}$. If $m=2$, then $b^{m-2} = 1\ne m$, so no solutions. If $m=3$, then $b=3$, so $(a,b) = (27,3)$. If $m=4$, then $b=2$, so $(a,b) = (16,2)$.

We claim that the only possible solutions are $(1,1)$, $(16,2)$, and $(27,3)$. Notice that for any prime $p$, if $\nu_p(a)=x$ and $\nu_p(b)=y$, we have that $xb^2=ay$. Taking this in two cases, if $b^2\geq{}a$, then we have that $y\geq{}x$ for all $p$, and if $b^2\leq{}a$, then we have that $y\leq{}x$ for all $p$, meaning that either $a\mid{}b$ or $b\mid{}a$. We now take this in cases. Case 1: $b=ak$ We have that $a^{a^2k^2}=a^ak^a$, so we end up with $k=a^{ak^2-1}$. We now prove that for all $(a,k)$ for $a,k>1$, we must have $k<a^{ak^2-1}$. We can fix $a$ and then proceed with induction on $k$. Notice that the base case of $k=2$ where $2<a^{4a-1}$ clearly works since $a>1$. Now notice that each time we add $1$ to $k$, the left side is multiplied by at most $2$, whereas the right side is multiplied by at least $a^a$, which is always at least $4$ since $a$ is an integer larger than $1$. Therefore no ordered pairs $(a,k)$ for $a,k>1$ work, and our only solution in this case is $(1,1)$. Case 2: $a=bk$ We have that ${bk}^{b^2}=bk$, which gives us that $k^b=b^{k-b}$. We want to now prove that for all $b>3$, we have that no integer $k$ satisfies the equation. Notice that we must have $b\mid{}k$, so we let $k=xb$. We then have $x=b^{x-2}$, and for $b>3$, we have that $x<b^{x-2}$, which can be proved with induction similar to the previous case. We now have that $b$ is either $1$, $2$, or $3$. Taking in cases and solving we see that our only solutions are now $(1,1)$, $(16,2)$, and $(27,3)$, and we are done.

Clearly, if $a=1\iff b=1$. When $a=2$ there are no solutions since $2^{b^2}>b^2$, and when $b=2$ this is $a^4=2^a$ which has $a=16$ as a solution. When $a=3$ this is $3^{b^2}=b^3$, which again has no solution since the left is much larger, and when $b=3$ this is $a^9=3^a$ which as $a=27$ as a solution. Noting these solutions, suppose $a,b\geq 4$ from now on. Claim: $b< a$. Suppose FTSOC that $b\geq a$. Then, from $$b^2=a \log_a(b),$$we have $$\frac{b^2}{\ln b}=\frac{a}{\ln a}.$$However, $\frac{x}{\ln x}$ is strictly increasing on $[4,\infty)$ so $$\frac{b^2}{\ln b}>\frac{b}{\ln b}\geq \frac{a}{\ln a},$$contradiction. Taking the base $a$ logarithm, we have $$b^2=a \log_a(b),$$so $b$ is a rational power of $a$. Thus, let $a=s^n$ and $b=s^m$ for positive integers $s,n,m$. Our equation then becomes $$s^{n\cdot s^{2m}}=s^{m\cdot s^n},$$so $$n\cdot s^{2m}=m\cdot s^n.$$This also means that $$s^{n-2m}=\frac{n}{m}.$$Let $n=s^cm$. Since $b<a$, we know that $c\geq 1$. Thus, $$s^{s^cm-2m}=s^c$$$$s^cm-2m=c$$$$m(s^c-2)=c.$$If $c=1$, then this is $$m(s-2)=1,$$so clearly $m=1$ and $s=3$, giving the solution $(27,3).$ If $c\geq 2$, then $$m(s^c-2)\geq s^c-2\geq 2^c-2\geq c,$$so the only possible case is all equalities, so $m=1$, $s=2$, and $c=2$, which gives the solution $(a,b)=(16,2)$ as expected earlier. Thus, the solutions are $(1,1),(16,2),(27,3)$ and we are done.

The only answers are $(1, 1)$, $(27, 3)$, and $(16, 2)$. Assume $a, b > 1$ henceforth. Let $d = \gcd(a, b)$ and let $a = dx$, $b = dy$, such that $$d^{b^2} \cdot x^{b^2} = d^a \cdot y^a.$$However, $x$ and $y$ cannot share any prime factors, so upon canceling the $d$'s, we must have $1 \in \{x, y\}$. In particular, $a \mid b$ or $b \mid a$. Suppose that $b \mid a$ first. Setting $a = db$, we have $$(db)^{b^2} = (b^d)^b \iff d^b = b^{d-b}.$$Using similar logic, we must have $b \mid d$ (as $d > b$). Hence, setting $d = rb$, $r = b^{r-2}$. Notice that for $b>5$, by monotonicity no solutions exist. So the only solutions here are $(b, r) = (3, 3) = (2, 4)$, which yield the two solutions above. Now set $a \mid b$. Setting $b = da$, we have $d = a^{d^2 a - 1}$. However, this clearly yields no solutions by monotonicity again. Thus, the set of solutions is given by the set above.

Solutions: $(1,1)$, $(16,2)$, $(27,3)$. Case 1: $b=a$. Then $a^2=a$ so $a=1$, and this leads to the solution $(1,1)$. Case 2: $b>a$. Then taking $\nu_p$ of both sides and rearranging we get $\frac{\nu_pb}{\nu_pa}=\frac{b^2}{a}$ (where we consider $\frac00$ to be equal to anything). Then $\nu_pa\le \nu_pb$ for all $p$, so $a\mid b\mid b^2$, so $\frac{b^2}a$ is an integer constant, say $c$; then $b=a^c$. Then $a^{a^{2c}}=a^{ac}$, so $a^{2c-1}=c$. Now, $a^c=b>a$ so $a>1$, and hence $c=a^{2c-1}\ge 2^{2c-1}$. As $b>a$, $c>1$, so $2c-1>c$, and hence $c>2^c$, which is not possible. Hence no solutions in this case. Case 3: $a>b$. Then since $a^{b^2}=b^a$ and we have $a>b$, we must have $b^2<a$, so $a>b^2$. Taking $\nu_p$ of both sides and rearranging we get $\frac{\nu_p a}{\nu_p b}=\frac{a}{b^2}$. Let the RHS in simplest form be $\frac cd$. Then for some integer $e\ge 2$, we must have $a=e^c$ and $b=e^d$. Then $e^{ce^{2d}}=e^{de^c}$. Then $ce^{2d}=de^c$. Now, we know $a>b^2$ so $c>2d$, and we get $\frac{c}{d}=e^{c-2d}$. Since the LHS is in simplest form we must have $d=1$ and so $c=e^{c-2}$. Now, since $e\ge 2$ we must have $c\ge 2^{c-2}$. Now, we know $c>2d=2$ so $c\ge 3$; then the possibilities are $c=3,4$. If $c=4$ then we have $4=e^2$ so $e=2$, which leads to the solution $(16,2)$. Otherwise $c=3$ so $3=e^{3-2}=e$, which leads to the solution $(27,3)$.

Answer: The only possible pairs are $(1, 1), (16, 2), (27, 3)$. We can check the case $\min(a, b) \le 2$ by hand, so we can assume $\min(a, b) \ge 3$. Since $a^b < a^{b^2} = b^a$, so $a^b < b^a$, or equivalently $\frac{a}{\log(a)} > \frac{b}{\log(b)}$. Since $\frac{x}{\log(x)}$ increases on $(e, \infty)$, hence $a > b$. Therefore $a^{b^2} = b^a < a^a$, so $b^2 < a$. Note that for all primes $p$ dividing both of $a$ and $b$, we have $\nu_p(a)b^2 = \nu_p(b)a > \nu_p(b)b^2$, so $\nu_p(a) > \nu_p(b)$. Thus $b \mid a$. Let $k = \frac{a}{b}$. Since $b^2 < a$, so $b < k$. The given condition can be rewritten as $k^b = b^{k-b}$. Since $k > b$, there exists a prime $p$ such that $\nu_p(k) > \nu_p(b)$. On the other hand, we have $\nu_p(k)b = \nu_p(b)(k-b)$, so we have $b < k - b$. Therefore for all primes $p$ dividing both of $b$ and $k$, we have $\nu_p(k)b = \nu_p(b)(k-b) > \nu_p(b)b$, thus $\nu_p(k) > \nu_p(b)$. Hence $b \mid k$. Let $k_1 = \frac{k}{b}$. The given condition can be rewritten as $k_1 = b^{k_1 - 2}$. Since $b \le 3$, we have $k_1 = b^{k_1 - 2} \ge 3^{k_1 - 2}$. Hence $2 < k_1 \le 3$, which forces $k_1 = 3$. Since $a = k_1b^2$, so we get $b = 3$ and $a = 27$. Therefore the solutions are $(1, 1), (16, 2), (27, 3)$. $\blacksquare$

Awful problem. First if $a=1,b=1$ and if $a=2$ we have no solutions and if $b=2$ we have $a=16.$ Now if $a,b\ge 3$ we see $a^{b^2}>a^b$ so we have $a^b<b^a$ which gives $a>b.$ Take logs to get $\log_a b=\frac{b^2}a.$ Now the right side is rational so the left side is rational and thus there must be some $p$ such that $p^A=a,p^B=b,\gcd(A,B)=1.$ We then have $p^{2B-A}=\frac BA$ so let $B=A\cdot p^{-k}$ and we get $2A=p^k(A-k)$ or $A=\frac{kp^k}{p^k-2}$ and since $\gcd(p^k-2,p^k)\le 2$ we must have $p^k-2 \mid 2k$ since $A$ is an integer. Now if $k\ge 4$ we have $p^k-2 \ge 2^k-2 > 2k$ so there are no solutions. If $k=2$ we have $p^2-2\mid 4$ so $p=2$ which gives $A=4,B=1$ which gives $(16,2)$ which we already have. If $k=3$ we have $p^3-2\mid 6$ so $p=2$ which gives $A=4,B=\frac12$ which is impossible. If $k=1$ we have $p-2\mid 2$ so $p=3,4.$ We see $p=4$ gives $A=2,B=\frac12$ which is impossible and $p=3$ gives $A=3,B=1$ which gives $(27,3)$ and thus all solutions are $(1,1),(16,2),(27,3).$

Taking $\log_b$ gives $b^2 \cdot log_b(a) = a \implies log_b(a) \in \mathbb Q \implies a \mid b$ or $b \mid a$. First, letting $a = bm$ gives $bm^{b^2} = b^{bm} \implies bm^b = b^m \implies m^b = b^{m-b}$. Using the $\log$ argument again we get that $b | m$ so setting $m = bp$ we get $bp^b = b^{bp - b} \implies bp = b^{p-1} \implies p = b^{p-2}$. For $b \geq 4$ this is impossible so we get that $b = 1, 2, 3$ which gives us the solutions $(a, b) = (1, 1), (16, 2), (27, 3)$. Then letting $b = am$ we get $a^{(am)^2} = (am)^a \implies a^{am^2 - 1} = m$ and this implies that $m = 1$ otherwise we get size issues. This then gives us the solution $(a, b) = (1, 1)$. So our solution set is $(1, 1), (16, 2), (27, 3)$.

Not bad. The only solutions are $(1,1), (16,2), (27,3)$. Note that $(1,1)$ is an obvious solution. Assume $a,b>1.$ Lemma : $v_p(n) < n$ for all $n \geq 2.$ If otherwise, then $p^n \mid n \implies p^n \leq n.$ This would imply $2^n \leq n,$ which clearly has no solutions. Hence, the result. It is clear that $a$ and $b$ have the same prime factors. We now show that $b^2 \mid a.$ First, we observe \[b^2 \nu_p(a) = a \nu_p(b) \implies b^2 \mid a \nu_p(b) \implies b^2 \leq a \nu_p(b) < ab.\] Thus, $a > b.$ Now, $a^{b^2} = b^a < a^a$ so $a > b^2$ as well. Thus, \[b^{2b^2} < a^{b^2} = b^a \implies a >2b^2.\] Now, as $b^2 \nu_p(a) = a \nu_p(b) \implies b^2\nu_p(a) > 2b^2\nu_p(b) \implies \nu_p(a) > \nu_p(b^2) $ for all $p \mid a,b.$ So $b^2 \mid a.$ Thus, $b^2 \nu_p(a) = a \nu_p(b) \implies \nu_p(a) = \nu_p(b)N$ for an integer $N \geq 3$, so $a = b^N.$ Assume $b \geq 4.$ From the original equation we get, \[b^N = Nb^2 \implies b^{N-2} = N \implies 3^{N-2} < b^{N-2} = N.\] not possible as $N \geq 3$. So, $b \in \{2,3\}$ which can be checked to give $a = \{16, 27\}$ by hand; giving us all the solutions.

The answer is $(a , b) = (1, 1), (16, 2), (27, 3)$. Let $\gcd(a, b) = d$ and $a = dp$, $b = dq$ where $\gcd(p, q) = 1$. Then we have $a^{b^2} = b^a \implies (dp)^{b^2} = (dq)^a \implies d^{b^2} p^{b^2} = d^a q^a$ Now condider the following cases: Case 1: $b^2 = a $ This gives $p = q$ but $\gcd(p, q) = 1$. So, $p = q = 1 $ and $a = b$.Then $a^2 = a$ so $a = 1$ and this leads to the solution $(a, b) = (1, 1)$. Case 2: $b^2 > a$ Here $d^{b^2 -a} p^{b^2} =q^a$ implies that $p$ divides $q^a$ but $ \gcd(p, q^a) = \gcd(p, q) = 1$, so, $p = 1$ and $a = d$, $b = a q$. So, $a^{b^2}= b^a \implies a^{(a q)^2}= (a q)^a \implies a^{a q^2 -1 } = q$. Clearly $a^{a q^2 -1 } > q$ for all $a > 1$ and for $a = 1$ we already have the solution. Case 3: $b^2 < a$ Using same argument as in case 2 gives $ q = 1 $ and $a = b p$. So, $a^{b^2}= b^a \implies (bp)^{b^2}= (b)^{b p} \implies p^b = b^{p - b}$. Notice that $p \geq 2b$. Let $ \gcd(p, b) = r $ and again using same argument as in case 2 gives $p = b r$. So, $ p^b = b^{p - b} \implies (br)^b = b^{br - b} \implies r = b^{r - 2}$. But $r < b^{r - 2}$ for $ r > 5 $, $b > 1$. So, for $r = 3$, we get $b = 3$ and $a = 27$ and for $r = 4$, we get $b = 2$ and $a = 16$. Thus, this case gives the solutions $(a , b) = (27, 3), (16, 2)$.

1997 IMO p5 We claim that the only answers are $(1,1) , (16,2)$ and $(27,3)$. By some $log$ argument we get that $b|a$ So we can sub in $a=xb$, so we get. \[(xb)^{b^2}=b^{xb}\]By similar argument we show that $b|x$ and hence this will force $b= 1,2,3$ Trying manually we get that desired solutions and we are done

Note that we have $\frac{2b}{a}=\frac{\nu_p(b)}{\nu_p(a)}$ thus we have that $a\mid b$ or $b\mid a$, if $a\mid b$ we get $a^{k^2(a-1)}=k$ where $b=ka$ and thus we must have $a=1$ and $k=1$ so $b=1$. If $b\mid a$, we get $k^b=b^{k-b}$ where $kb=a$, thus we get $b\mid k$ as $k-b\geq 0$, so let $mb=k$ we get $mb=b^{m-2}$ and it suffices to check $m\in \{1, 2, 3, 4\}$ and from checking the cases we reach $2, 16$ and $3, 27$ as the other solutions.

Let $b = a^{r}.$ Then, taking $\log_{a}$ of both sides, we find that $ra = b^{2},$ so $r$ is rational. Thus, $a^{2r-1}=r.$ So, if $r = \frac{m}{n},$ we have that $m = 1$ or $n=1.$ \newline \newline If $m=1,$ then $$a^{\frac{2-n}{n}}=\frac{1}{n}.$$Thus, since $\gcd(2-n,n) = \gcd(2,n) \mid 2,$ we have that $n^{\frac{2}{n}}$ is an integer. \begin{claim} If $n^{\frac{2}{n}}$ is an integer, then $n = 1,2, 4.$ \end{claim} \begin{proof} If $n^{\frac{2}{n}} = 1,$ then $n=1.$ Otherwise, $n^{\frac{2}{n}} \ge 2.$ We claim that $2^{n}>n^{2},$ and we will revert to induction to prove it. The base case amounts to $32 > 25.$ For the inductive step, assume that $2^{n}>n^{2}.$ Then, $$2^{n+1}=2 \cdot 2^{n} > \frac{(n+1)^{2} }{n^{2}} \cdot 2^{n}>\frac{(n+1)^{2}}{n^{2}} \cdot n^{2} = (n+1)^{2},$$as desired. Thus, this is true. So, $n^{\frac{2}{n}} < 2$ for $n \ge 5.$ So, $n=1,2,3,4.$ Obviously, $n \neq 3,$ so $n=1,2,4$ are the only values that work. \end{proof} Plugging these values of $n$ in, we find that if $n = 1,$ then $(a,b) = (1,1),$ if $n=4,$ then $(a,b)=(16,2),$ but if $n=2,$ then there are no solutions. \newline \newline If $n = 1,$ then $m^{\frac{m}{2m-1}}$ is an integer. Now $\gcd(m,2m-1)=1,$ by the Euclidean Algorithm, so $m^{\frac{1}{2m-1}}$ is an integer. Obviously $m < 2^{2m-1},$ so $m=1.$ Hence we get that $(a,b) = (1,1).$

seems like a 0 mohs but took so long wth

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