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Source: IMO 1997, Problem 5, IMO Shortlist 1997, Q17

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hxtung

#1 h Oct 13, 2003, 10:11 AM • 15 Y

Y by ShahinH, ahmedosama, Davi-8191, hashtagmath, HarryPotterAnis, mathematicsy, Adventure10, justJen, HWenslawski, megarnie, Mango247, and 4 other users

Find all pairs $(a,b)$ of positive integers that satisfy the equation: $a^{b^2} = b^a$ .

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probability1.01

2743 posts

probability1.01

#2 h Nov 7, 2005, 3:46 PM • 13 Y

Y by ShahinH, Math1331Math, HarryPotterAnis, Bumblebee60, Adventure10, HWenslawski, Gaunter_O_Dim_of_math, MarioLuigi8972, and 5 other users

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.

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Nanas

#3 h Sep 10, 2013, 10:49 PM • 5 Y

Y by spin8, opptoinfinity, HarryPotterAnis, Adventure10, Mango247

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 $(a,b) \in \{(1,1),(2^4,2),(3^3,3)\}$

This post has been edited 4 times. Last edited by Nanas, Feb 27, 2014, 8:04 PM

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#4 h Sep 10, 2013, 11:00 PM • 6 Y

Y by Nanas, Wavefunction, HarryPotterAnis, amuthup, Adventure10, Mango247

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)

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#5 h Aug 26, 2017, 11:57 PM • 1 Y

Y by Adventure10

@Nanas
Why m is equal to 1

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LoveMaths26102003

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LoveMaths26102003

#6 h Dec 12, 2017, 3:25 PM • 1 Y

Y by Adventure10

@oiviluhatef can you express the technique in detail??

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779 posts

#7 h Mar 22, 2020, 11:52 PM

Y by

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?

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#8 h Mar 23, 2020, 12:42 AM • 3 Y

Y by Mango247, Mango247, Mango247

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

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386 posts

#9 h Mar 24, 2020, 4:28 PM • 1 Y

Y by Mango247

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:

$\boxed{(a,b) \in \{(1,1);(16,2);(27,3) \}}$

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4 posts

#10 h Jan 29, 2021, 12:42 PM • 1 Y

Y by Mango247

Does this equation have negative roots?

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988 posts

test20

#11 h Jan 29, 2021, 2:35 PM

Y by

Birdews wrote:

Does this equation have negative roots?

What about $a=b=-1$ ?

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6858 posts

#12 h Mar 18, 2021, 10:18 PM • 5 Y

Y by HamstPan38825, bluelinfish, MrOreoJuice, Mango247, Mango247

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.

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#14 h Sep 22, 2021, 7:27 PM

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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.

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351 posts

#15 h Oct 8, 2021, 5:48 PM • 1 Y

Y by Bradygho

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 $(27,3)$ .

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#16 h Oct 28, 2021, 12:14 PM

Y by

Beautiful

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#17 h Oct 29, 2021, 10:46 PM • 1 Y

Y by centslordm

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. $\blacksquare$

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#18 h Jan 20, 2022, 9:02 PM • 1 Y

Y by centslordm

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.

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#19 h Jan 28, 2022, 8:43 PM • 1 Y

Y by ZHEKSHEN

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 $d-2\geq 1$

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656 posts

#20 h Feb 6, 2022, 10:58 PM

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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. $\blacksquare$

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#21 h Jan 15, 2023, 11:45 AM

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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) ?

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#22 h Jan 16, 2023, 4:11 PM

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Anyone ?

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#23 h Jan 17, 2023, 6:29 PM • 2 Y

Y by Mango247, Mango247

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 $(27,3)$

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5533 posts

#24 h Jan 18, 2023, 2:41 AM • 3 Y

Y by Mango247, Mango247, Mango247

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 $(a,b) = (16,2)$ .

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#25 h Mar 15, 2023, 11:26 PM • 5 Y

Y by mahaler, Significant, happymathEZ, megarnie, mulberrykid

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.

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#26 h Aug 3, 2023, 8:19 PM

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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.

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#27 h Dec 2, 2023, 6:00 PM

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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.

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#28 h Dec 2, 2023, 6:50 PM

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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 $(27,3)$ .

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#29 h Feb 6, 2024, 5:53 AM

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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 $(1, 1), (16, 2), (27, 3)$ . $\blacksquare$

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OronSH

#30 h Feb 9, 2024, 2:10 AM • 1 Y

Y by megarnie

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 $(1,1),(16,2),(27,3).$

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#31 h Mar 29, 2024, 6:16 PM

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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 $(1, 1), (16, 2), (27, 3)$ .

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#32 h May 5, 2024, 7:55 AM • 1 Y

Y by bissonter

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.

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#33 h May 6, 2024, 3:53 PM

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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 $(a , b) = (27, 3), (16, 2)$ .

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#34 h Aug 27, 2024, 8:14 AM

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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

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#35 h Sep 21, 2024, 8:09 AM

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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.

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#36 h Oct 4, 2024, 9:40 PM

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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 $(a,b) = (1,1).$

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#37 h Jan 10, 2025, 5:57 PM

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seems like a 0 mohs but took so long wth

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#38 h Yesterday at 5:42 PM

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Our solutions are $\boxed{(1,1), (16,2), (27,3)}$ . The solution $a=b=1$ is obvious, so we suppose $a,b > 1$ for the remainder of our solution.

Note $a$ , $b$ must both be powers of some integer, so define $(a,b) = (k^x, k^y)$ with $k$ maximal (i.e. $\gcd(m,n) = 1$ ). Comparing exponents, we get
$xb^2 = ya \implies k^{2y-x} = \frac yx.$
Since $m$ , $n$ are coprime, at least one must be equal to 1, from which simple casework and size bounding finishes. $\blacksquare$

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