# Equality of power towers : $a\uparrow\uparrow m=b\uparrow \uparrow n$

Suppose, $$a,m,b,n$$ are natural numbers greater than $$1$$.

If we have $$a\uparrow\uparrow m=b\uparrow\uparrow n$$

can we conclude $$a=b$$ and $$m=n$$ ?

$$a\uparrow \uparrow m$$ is a powertower of $$m$$ $$a's$$ and $$b\uparrow\uparrow n$$ a power tower of $$n$$ $$b's$$.

I proved $$b^b\ne a^{a^a}$$ for natural numbers $$a,b>1$$.

This is my proof :

Set $$s:=\frac{ln(b)}{ln(a)}$$. So, we have $$a^s=b$$. Therefore, we have

$$a^{a^a}=(a^s)^{(a^s)}=a^{sa^s}$$ , so $$a^a=sa^s=sb$$, we can conclude that $$s$$ is rational.

If $$s\ge a$$, then $$sa^s\ge aa^a>a^a$$. If $$s\le a-1$$, then $$sa^s=(a-1)a^{a-1}

Hence, we can conclude $$a-1 implying that $$s$$ is not an integer.

Finally, assume $$s=\frac{m}{n}$$ with coprime $$m$$ and $$n$$. Then, we have

$$(\frac{m}{n})^na^{s\ n}=a^{an}$$, implying $$(\frac{m}{n})^n=a^{an-m}$$

Because of $$a>s$$ we have $$an-m>sn-m=0$$. Therefore, the left side of the last equation is not an integer, while the right side is an integer. This contradiction finally proves the claim.

Questions:

Is this proof correct ?

Is there an easier proof that $$a^{a^a}=b^b$$ cannot hold for natural numbers $$a,b>1$$ ? (For example, using the $$p$$-adic value of $$a^{a^a}$$ and $$b^b$$) ?

Is there a proof that we can extend to show the more general initial claim ?

• I found no error; upvote because I like the resoning with the s-value, which surprised me initially but it comes out that the arguing about its properties is clear and has nice ideas Commented May 12, 2016 at 6:53

One can prove your initial claim by proving the following more general statement:

Theorem. $a^{a^b} = c^{c^{c^d}}$ with $a > c \ge 2$ and $b,d \ge 1$ has no solutions.

Proof. Write $a^z = c^y$ with $\text{gcd}(z,y) = 1$. Then $a^{\frac{1}{y}} = c^{\frac{1}{z}} = x$ is an integer, and $a = x^y, c = x^z$. So

$$x^{y x^{yb}} = x^{z x^{z x^{zd}}}$$

$$y x ^{yb} = z x^{z x^{zd}}$$

hence $\frac{y}{z}$ is a positive power of $x$; let $y = x^w$. Then

$$z x^w x^{zx^w b} = zx^{zx^{zd}}$$ $$w + zx^w b = zx^{zd}$$

But, since $w < zx^w$, $w + zx^w b$ lies between $zx^w b$ and $zx^w(b+1)$, and therefore cannot be of the form $zx^{zd}$.

• Nice proof, but I will need some time to completely follow it. Commented May 16, 2016 at 17:47