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

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Presumably, $b = x^x$ should be $b = x^a$. – JavaMan Apr 3 '11 at 4:13
Is it necessary that a and b are coprime? – awllower Apr 3 '11 at 4:14
@DJC: Since $a=x$, it doesn't matter. – joriki Apr 3 '11 at 5:28
What is the $y$ for? – Kea Apr 3 '11 at 6:14
@Jan: Sorry, I meant $b$. Edited now. I have seen a solution, but it is, IMHO, an ugly one. I will post it if needed. :) – Amir Hossein Apr 4 '11 at 18:01
up vote 11 down vote accepted

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 $b=x^a$.

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There seems to be a problem with the initial part: $p^{n_b}\nmid n_b$ only holds of $n_b\ne 0$. So really we have $n_b > n_a$ or $n_b=0$. (Otherwise we would be done much more quickly since $n_b$ would have to have every prime number as a factor!) Similarly when you conclude that $p^{n_a}\le n_b$ you are assuming $n_b$ is positive. Perhaps you could just restrict attention to primes dividing $b$ but then it's not clear how you get that $a \mid n_b$. – Harry Altman Apr 5 '11 at 23:52
@Harry: Thanks! I'll see whether I can fix that... – joriki Apr 6 '11 at 4:44
@Harry: I edited to treat the case $n_b=0$ -- hope everything's OK now? – joriki Apr 6 '11 at 5:19
very nice, thanks! – Amir Hossein Apr 6 '11 at 15:50

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