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Solve $x^{y^2}=y^x$ which $x,y\in\mathbb{N}$.

I can observe that $(x,y)$ can be $(1,1)$, then I don't know how to carry on. Please help. Thank you.

p.s. I wonder if there's solution without using mod.

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Two solutions : $(16,2)$, $(27,3)$. – Spenser Oct 17 '12 at 13:18
@ Spenser: Steps please? – ᴊ ᴀ s ᴏ ɴ Oct 17 '12 at 13:25
Sorry, I got this only by looking at the equation and trying a few things. Having some solutions could help to find other ones or a general pattern though. – Spenser Oct 17 '12 at 13:38
$y=p$(prime) and $x=p^a,\implies ap^2=p^a\implies a=p^{a-2}$, so $a$ is power of $p$, $p^k=p^{p^k-2}\implies p^k=k+2$, $if k=1,p=2; k=2,p=3$ and $p^k>k+2$ for $p>3$ – lab bhattacharjee Oct 17 '12 at 13:39
@Spenser Thanks, your two solution values led me to the right hypothesis that necessarily $y^2 \mid x$. – Lord_Farin Oct 17 '12 at 14:02
up vote 7 down vote accepted

By the argument I have added below, we establish $x = my^2$ if $x > y$, for $m \in \Bbb N$.

Plugging in gives: $(my^2)^{y^2} = y^{my^2} \iff my^2 = y^m \iff m = y^{m-2}$. Demanding that $y\ge 2$ implies that $m \le 4$.


  • $m=1,2$: Impossible since $y \in \Bbb N$, $y^0 = 1$.
  • $m=3$: Plugging in gives $3 = y$, yielding the second solution by @Spenser.
  • $m=4$: Plugging in gives $4 = y^2$, yielding the first solution by @Spenser.

EDIT: The argument that $x = my^2$ if $x > y$:

Suppose that $x > y$. We have

$$x = \log_y y^x = \log_y x^{y^2} = y^2 \log_y x \implies \frac x{y^2} > 0$$

so that $x > y^2$. Now for any prime $p$, we have:

$$x \operatorname{ord}_p y = y^2 \operatorname{ord}_p x$$

Therefore, if $\operatorname{ord}_p y > 0$, it must be that $\operatorname{ord}_p x > \operatorname{ord}_p y$ to be able to have equality. That is, $y \mid x$.

So put $x = ny$ with $n \in \Bbb N$. Then $n > y$ since $x > y^2$, and:

$$(ny)^{y^2}=y^{ny} \iff n^y = y^{n-2}$$

Resorting again to the $\operatorname{ord}_p$ trick, we find similarly that $\operatorname{ord}_p n > \operatorname{ord}_p y$, i.e. $y \mid n$. Therefore, $x = my^2$ for some $m \in \Bbb N, m \ge 2$.

EDIT 2: The argument that $x \ge y$ for any solution, thus completing the proof that all solutions have been found (since $x = y$ only gives $(1,1)$ as solution):

Suppose that $x < y$. Then since $x < y^2$, employing

$$x \operatorname{ord}_p y = y^2 \operatorname{ord}_p x$$

again yields $x \mid y$, say $y = nx$. Then we obtain, plugging this into the equation:

$$x^{(nx)^2} = (nx)^x \iff x^{n^2x} = nx \iff n = x^{n^2x-1}$$

Because $x \ge 2$, this has no solutions for $n \in \Bbb N$.

It thus follows that the solution space is given by:

$$\left\{(x,y) \in \Bbb N^2: x^{y^2}=y^x \right\} = \{(1,1), (16,2), (27,3)\}$$

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You can rewrite the equation as $x = y^{xy^{-2}}$, plugging this into itself gives $x = y^{y^{xy^{-2}-2}}$, and by taking logs (and supposing $y > 1$) we get $y^{xy^{-2}-2} = xy^{-2}$. Finally, raising to the $y^2$th power, we get the equality $y^{x-2y^2} = (xy^{-2})^{y^2}$

If $x > 2y^2$, the LHS is an integer, the RHS is a rational to an integer exponent. This forces the rational on the right to be an integer, and so $x$ has to be a multiple of $y^2$. So $x = my^2$ for some $m > 2$. Then the equation simplifies to $y = m^{1/(m-2)}$ which has the two solutions for $m=3$ and $m=4$

If $x \le 2y^2$ we can take the inverses and now it is $y^2$ that has to be a multiple of $x$. If $y^2 = mx$, we get $(x^{2m})^x = (mx)^x$ thus $x = m^{1/(2m-1)} < 2$, and this is possible only for $x=y=m=1$.

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That's nice; I didn't give it second thought once I saw the argument with primes work. – Lord_Farin Oct 17 '12 at 14:19

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