Let $a,x,b$ be positive integers satisfying $x^{a+b} = a^b \cdot b$. How can I prove that $a=x$ and $b=x^x$?

  • $\begingroup$ Is it necessary that a and b are coprime? $\endgroup$ – awllower Apr 3 '11 at 4:14
  • $\begingroup$ What about $x=b=0$ and $a \not = 0$ $\endgroup$ – picakhu Apr 6 '11 at 6:13
  • $\begingroup$ @picakhu: It says "positive integers". $\endgroup$ – joriki Apr 6 '11 at 6:37
  • $\begingroup$ @picakhu: This was going to be a comment on the answer that you just deleted. I guess you've found the error in the meantime, but two more things: I was wrong about the bug in Wolfram Alpha: I think the zero result is indeed correct if you change the sign you got wrong (or equivalently swap $a/x$ back to $x/a$) -- but of course then the rest of your proof doesn't work. Note also that we're trying to prove $a=x$, so you can only divide by $\ln(x/a)$ as part of an indirect proof assuming $a\neq x$. $\endgroup$ – joriki Apr 6 '11 at 14:04

The equation must be satisfied for each prime individually; that is, if $p$ is a prime factor of any of $a$, $b$ and $x$ and we denote the number of factors of $p$ in $a$, $b$ and $x$ by $n_a$, $n_b$ and $n_x$, respectively, we must have


[Edit: Thanks to Harry for pointing out that I forgot to treat the case $n_b=0$. In this case, $an_x=b(n_a-n_x)$, and so $p^{n_a}\mid n_a-n_x$ (since $b$ contains no factors of $p$). But $n_a\ge n_x$, since $(a+b)n_x=bn_a$, and so either $n_a=n_x$, which would imply $n_a=n_x=n_b=0$, or $n_a>n_x$, and thus $n_a\ge n_a-n_x\ge p^{n_a}$, which is impossible. Thus $n_b\neq0$.]

Now let $q$ be any prime factor (not necessarily distinct from $p$) in any of $a$, $b$ and $x$, and denote the number of factors of $q$ in $a$ and $b$ by $m_a$ and $m_b$, respectively. Then $q^{m_a}| a$ and $q^{m_b}| b$, and hence each term in the equation except for $n_b$ is divisible by $q^{\min(m_a,m_b)}$; thus $n_b$ is, too. In particular, $n_b$ is divisible by $p^{\min(n_a,n_b)}$. But $p^{n_b} \nmid n_b$, since $p^{n_b}>n_b$, and hence $n_b>n_a$. Since $p$ and $q$ are arbitrary, this implies $m_b>m_a$, and thus $q^{m_a} | n_b$. In particular $p^{n_a} | n_b$, and thus $p^{n_a}\le n_b$. Also, since $q^{m_a} | n_b$ for all prime factors $q$ of $a$, we have $a|n_b$. Thus we can write (1) as


To show that both sides of this equation are in fact zero, we can again consider factors of $p$. The right-hand side contains at least $n_b$ factors of $p$, and $a$ contains only $n_a$, so

$$p^{n_b-n_a}\mid n_x-\frac{n_b}{a}\;,$$

and thus if this difference is not zero, we must have

$$p^{n_b-n_a} \le \left\lvert n_x-\frac{n_b}{a}\right\rvert\;.$$

Considering first the case $n_x<\frac{n_b}{a}$, it follows that

$$ p^{n_b-n_a} \le \left\lvert n_x-\frac{n_b}{a}\right\rvert = \frac{n_b}{a}-n_x \le \frac{n_b}{a} \le \frac{n_b}{p^{n_a}}\;, $$

and thus $p^{n_b}\le n_b$, which is impossible. Considering instead the case $n_x>\frac{n_b}{a}$, from (2) this implies $n_a>n_x$, and thus

$$ p^{n_b-n_a} \le \left\lvert n_x-\frac{n_b}{a}\right\rvert = n_x-\frac{n_b}{a} \le n_x < n_a < p^{n_a}\;, $$

that is, $p^{n_b}<p^{2n_a}$ and thus $n_b<2n_a$. But we had $p^{n_a}\le n_b$, and thus $p^{n_a}<2n_a$, which is again impossible.

It follows that both sides of (2) vanish, and thus $n_a=n_x$ and $n_b=an_x$ for all primes $p$, and this implies $a=x$ and $b=x^a$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.