# Find all solutions in $\mathbb{N}$ to $a^a=a^b+b^a$

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.

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

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

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

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

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

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

• My bad, I didn't understand you were aiming for a contradiction. Very nice proof! – 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. $$\blacksquare$$

• Very nice use of induction and the lemma. (+1) – 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.

• 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. – URL Jan 5 '20 at 20:04
• I agree but I was just so pleased when I saw an FLT equation I had to use it! – S. Dolan Jan 5 '20 at 20:15
• If you really wanted to conclude with something overpowered, you could've just cited Catalan/Mihăilescu ;) – URL Jan 5 '20 at 20:18
• 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. – 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, $$a/b>2$$

• 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. – URL Jan 5 '20 at 20:16
• yeah I made a mistake there. edited my comment. – SL_MathGuy Jan 5 '20 at 20:20
• 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. – URL Jan 5 '20 at 20:31
• 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. – SL_MathGuy Jan 5 '20 at 20:35
• "It is safe to argue" is very different from "we can prove that". This is still a gap in your argument. – URL Jan 5 '20 at 20:40