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I have the following question:

If $a$ and $b$ are positive integers such that $b > a$, when will $a^b = b^a$?

I could not solve it, so I looked at the answer:

If $a$ and $b$ are positive integers such that $b > a$, $a^b = b^a$ when $a = 2$ and $b = 4$.

It is not easy to prove that this is the only solution; students should use guess and check to find a solution.

The answer provides no proof or insight into the method for finding the solution, apart from a vague mention of "guess and check".

I wish to find the (algebraic, if possible) proof for the question. To sum up:

Given that $a > 0$, $b > 0$ and $b > a$, prove that the only solution of $a^b = b^a$ is that $a = 2$ and $b = 4$.

Aryan
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    Does this answer your question? Find all solutions of $a^b = b^a$ – scoopfaze Jan 31 '20 at 20:04
  • These might help: https://math.stackexchange.com/questions/871169/find-all-solutions-of-ab-ba

    https://math.stackexchange.com/questions/793755/given-that-xy-yx-what-could-x-and-y-be

     – scoopfaze Jan 31 '20 at 20:04
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    Aryan, this site has really many duplicates of your question (it is a very popular one). You should search next time before posting. – Dietrich Burde Jan 31 '20 at 20:05
  • But none of them actually prove the solution. – Aryan Jan 31 '20 at 20:06
  • Not true, the answers here are very detailed and precise and got many upvotes. I suspect you haven't looked for it, right? And there are so many more proofs at this site for this. – Dietrich Burde Jan 31 '20 at 20:07
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    It is known that all rational solutions $(a,b)$ with $a<b$ to $a^b=b^a$ takes the form $$a=\left(1+\frac{1}{n}\right)^n$$ and $$b=\left(1+\frac{1}{n}\right)^{n+1},$$ where $n$ is a positive integer. From this result, it follows immediately that the only integer solution is $(a,b)=(2,4)$ corresponding to $n=1$. – Batominovski Jan 31 '20 at 23:47

2 Answers2

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Suppose $a>b$. Since $a^b=b^a$, any prime that divides $a$ must divide the right-hand side, so it divides $b$. Similarly, any prime dividing $b$ divides $a$, so $a$ and $b$ have the same prime factors. Say $$ \begin{align}a&=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}\\ b&=p_1^{b_1}p_2^{b_2}\cdots p_k^{b_k}\end{align}$$ for primes $P_i$ and positive integers $a_i,b_i$.

We have $$ p_1^{ba_1}\cdots p_k^{ba_k}=a^b=b^a=p_1^{ab_1}\cdots p_k^{ab_k}$$ By the fundamental theorem of arithmetic, $$ba_i=ab_i,\ i=1,\dots,k$$

Since $\frac{a}b>1$, we have $a_1>b_i$ for $i=1,\dots,k$ and so $\frac ab=p_1^{a_1-b_1}\cdots p_1^{a_k-b_k}$ is an integer, say $a=nb$.

Now we have $(nb)^b= b^{nb}\implies \sqrt[n-1]{n}=b$. Since $b$ is an integer, we must have $n=1$ and then $b=2, a=4$.

saulspatz
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Let $a,b$ be two positive integers such that $b > a$.

The equation $a^b =b^a$ would imply that $b= a^2.$ And since $b > a$, then $a > 1.$

So we have the following:

$a^{a^2} =b^{2a}$

Applying natural logarith to both sides: $ln(a^{a^2}) =ln(b^{2a})$

Applying the natural logarith property: ${a^2}*ln(a) =2a * ln(a)$

Dividing both sides by $ln(a).$ Here there is no problem doing this because we deduced that $a > 1.$

${a^2} =2a, \rightarrow a=2.$

And since $b= a^2, b= 4.$

StoneMather
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