Problem: Find all positive integer solutions to the equation $a^a=a^b+b^a$.

Attempt at a solution: I first acknowledged that $a > b$, since if $a \le b$, then $a^a=a^b+b^a\ge a^a+a^a=2a^a$, which isn't true for any positive integer $a$. Then I used $a-b=d$, which, of course means $a=b+d$, and I substituted that into the original equation. I got $(d+q)^d[(d+q)^d-1]=q^dq^q$. I'm stuck here. Any help would be appreciated.

  • $\begingroup$ Another thing to notice is that $a^b | b^a$ $\endgroup$ – SL_MathGuy Jan 5 '20 at 19:40
  • $\begingroup$ of course I screwed up my earlier (now deleted) comment $(b+d)^b(b+d)^d=(b+d)^b+b^{b+d}$ $\endgroup$ – user645636 Jan 5 '20 at 19:47
  • $\begingroup$ @RoddyMacPhee $a^b\mid b^a$ does not imply $a\mid b$. The most you can immediately conclude is $\text{rad}(a)\mid b$. $\endgroup$ – URL Jan 5 '20 at 19:48

As you noted, $a>b\geq 1$. Then

$$0=a^b+b^a-a^a\leq a^{a-1}+(a-1)^a-a^a$$

$$0\leq \left(1-\frac{1}{a}\right)^a+\frac{1}{a}-1=\left(1-\frac{1}{a}\right)^a-\left(1-\frac{1}{a}\right)$$


As this is a contradiction, we conclude there are no positive integer solutions.

  • $\begingroup$ My bad, I didn't understand you were aiming for a contradiction. Very nice proof! $\endgroup$ – URL Jan 5 '20 at 20:26

We already know that $a>b$. With this, we can prove that $$a^a>a^b+b^a,$$ thus proving that no solutions exist. We first prove a little lemma, and then we use induction on $a$.

Lemma: If $a>b$, $$a^{b+1}>(a+1)^b.$$

Proof: If $a=2$, then $b$ is necessarily equal to $1$, and the inequality holds, as $4>3$. Otherwise, we have $$a>e>\left(1+\frac1a\right)^a>\left(1+\frac1a\right)^b\Rightarrow$$ $$a^{b+1}>(a+1)^b,$$ as wanted. Here, $e\approx2.71828$ is Euler’s constant. $\square$

With this, we begin our proof.

If $a=b+1$, the base case, we have $$(b+1)^b>b^b\Rightarrow$$ $$b\cdot(b+1)^b>b^{b+1}\Rightarrow$$ $$(b+1)^{b+1}>(b+1)^b+b^{b+1}.$$ Now, assuming $a^a>a^b+b^a$ for some $a>b$, we have $$(a+1)^{a+1}>a^{a+1}>a^{b+1}+a b^a>(a+1)^b+b^{a+1},$$ by our induction hypothesis and our lemma. This completes our proof. $\blacksquare$

  • $\begingroup$ Very nice use of induction and the lemma. (+1) $\endgroup$ – S. Dolan Jan 5 '20 at 20:13

$a^a>b^a$ and so $a>b$. Let $a=b+d$.

Consider any prime $p$ dividing $a$ and let the maximum powers of the prime $p$ dividing $a$ and $b$ be $p^k$ and $p^l$, respectively. Then comparing the powers of $p$ dividing each side of $$a^b(a^d-1)=b^{b+d}$$ we obtain $bk=(b+d)l$.

Let $\frac{b}{b+d}=\frac{u}{v}$, where $u$ and $v$ are coprime. Then there is a positive integer $t$ such that $a=tv,b=tu$.

Also, there is a positive integer $s$ such that $k=sv,l=su$. Then $a$ is a $v$th power and so there is a positive integer $N$ such that $a=N^v$ and $b=N^uM$, where $N$ and $M$ are coprime.

Then the original equation cancels down to $$N^{vt(v-u)}-1=M^{tv}. $$ By FLT we have $tv\le2$ i.e $a\le2$. There are no solutions.

  • $\begingroup$ Two small pet peeves. For one, you really didn’t use any property of $d$, and your manipulations would’ve been slightly cleaner just using $b-a$. Also, you can conclude $vt=1$ without using something as nuclear as FLT just by noticing that two positive powers of $vt$ must be separated by at least $2^{vt}-1^{vt}$. Great solution otherwise. $\endgroup$ – URL Jan 5 '20 at 20:04
  • $\begingroup$ I agree but I was just so pleased when I saw an FLT equation I had to use it! $\endgroup$ – S. Dolan Jan 5 '20 at 20:15
  • $\begingroup$ If you really wanted to conclude with something overpowered, you could've just cited Catalan/Mihăilescu ;) $\endgroup$ – URL Jan 5 '20 at 20:18
  • $\begingroup$ It does't come to my mind as readily as FLT. (In fact I still think of it as a conjecture - showing my age I guess). I've improved the unnecessary use of $d$ , thanks. $\endgroup$ – S. Dolan Jan 5 '20 at 20:27

____________Edited version______________

Note that $a>b$. Dividing everything by $a^b$, we obtain,

$a^{a-b}=1+\frac{b^a}{a^b}$. You can see the term on the L.H.S must be an integer. Hence, $b^a$ must be divisible by $a^b$.So, $b^a=ka^b$ for $k \in \mathbb{N}$ & $k>1$. Substituting this in the orginial equation, we get,

$k (\frac{a}{b})^a= k+1$. So, $(1+\frac{1}{k})^{1/a} = \frac{a}{b}$.Clearly, $(1+\frac{1}{k})^{1/a}<2$ but $a/b>2$. This is a contradiction as there exists no such positive integer $k$ that satisifies this relation.

Hence,there exist no solutions for this equation.

Proof of $a/b >2$.

Suppose $a/b≤2$. What are the possible values of $a$ & $b$ that satisfy this inequality?(we can assume $a=2b$ since $a≠b$). i.e $k=2$. But then, we obtain $(a/b)^a=2^a=3/2$ (by $(1+\frac{1}{k}) = (\frac{a}{b})^{a}$). This is a contradiction. Hence, $a/b>2$

  • $\begingroup$ I don’t understand your last step. Why can’t there exist an integer $k$ such that $k\left(\frac ab\right)^a=k+1$? This seems like an enormous stretch. $\endgroup$ – URL Jan 5 '20 at 20:16
  • $\begingroup$ yeah I made a mistake there. edited my comment. $\endgroup$ – SL_MathGuy Jan 5 '20 at 20:20
  • $\begingroup$ I still don't get it. I suppose that by "$a>>b$", you mean $\frac ab>2$. Where does that come from? This seems too informal to be a complete argument. $\endgroup$ – URL Jan 5 '20 at 20:31
  • $\begingroup$ Let's consider the case $k=1$. Then $b^a=a^b$. The only solutions for this are $a=4$ & $b=2$ ($a=2$ & $b=4$ are left out). But, these values of $a$ & $b$ don't satisfy our original expression. That's why I assumed $k>1$. So, it's safe to argue that $a/b >2$. Precisely, $a>>b$, if we assume there exist solutions of this equation. $\endgroup$ – SL_MathGuy Jan 5 '20 at 20:35
  • $\begingroup$ "It is safe to argue" is very different from "we can prove that". This is still a gap in your argument. $\endgroup$ – URL Jan 5 '20 at 20:40

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