# conditional equation

Find all pairs $(x, y)$ of integers such that $x \ge 1$ and $y \ge 1$ and $x^{y^2} = y^x$.

work done: if $d = \gcd (x, y)$. then $x = du$ and $y = dv \implies \gcd (u, v) = 1$ and the equation becomes $(du)^dv^2 = (dv)^u$ if $dv^2 = u$ then $u = 1 = v$, $d = 1$ and hence $x = 1 = y \implies (1, 1)$ is one of the solution.

Now my question is, if $dv^2 > u$ and $dv^2 < u$, what are the solutions and how to conclude?

Find all pairs of $(x, y)$ such that $x \ge 1$ and $y \ge 1$ of $x ^{(y^2)} = y^x$.

work done: if $d = \gcd (x, y)$. then $x = du$ and $y = dv\implies \gcd (u, v) = 1$ and the equation becomes $(du)^{(dv^2)} = (dv)^u$ if $dv^2 = u$ then $u = 1 = v$, $d = 1$ and hence $x = 1 = y\implies (1, 1)$ is one of the solutions.

Now my question is, if $dv^2 > u$ and $dv^2 < u$, what are the solutions and how to conclude?

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gandhi: I've tried to improve LaTeX formatting, but I left the original version too - to be sure I did not change something. (If you're satisfied with the LaTeXed version, you can put the original one away.) BTW I think that at this place in your post: "(du)^d$v^2$" your missing $d^2$ instead of $d$ in the exponent. –  Martin Sleziak Apr 8 '12 at 7:32
@Martin: I believe that gandhi divided both exponents by $d$, getting $(du)^{dv^2}=(dv)^u$ from $(du)^{d^2v^2}=(dv)^{du}$. –  Brian M. Scott Apr 8 '12 at 7:39
No, Martin Sleziak is very much correct. whatever Martin is edited is correct. Thank you –  gandhi Apr 8 '12 at 8:18
Added the requirement that $x,y$ be integers. –  GEdgar Apr 9 '12 at 18:06

This solves the $(x,y)$ problem but doesn't follow your approach.

If $x=1$ or $y=1$ then $(x,y)=(1,1)$.

If $x>1$ and $x^{(y^2)}=y^x$ then $y^2 = x\log_x y$ and both sides can be integers only if $y=w^p$ and $x=w^q$ for positive integers $w,p,q$.

Then $y^2=x\log_x y \implies w^{2p-q} = \frac{p}{q}$ and since $w$ is an integer, either $q=1, 2p-q\ge 0$ or $p=1,2p-q\le 0$.

In the first case $y=x^p$, $x^{2p}=xp$, so $2p-1=\log_x p$. This has no solutions in $\mathbb{R}$ for $x\ge 2$ and $p>\frac{1}{2}$ so no integer solutions to our problem.

In the second case $x=y^q$, $y^2=y^q/q$, so $\log_y q = q-2$. For $y\ge 2$ this has one solution for $q\ge 1$ in $\mathbb{R}$. If $y=2$ then $q=4,x=16$, if $y=3$ then $q=3,x=27$, and if $y>3$ then $2<q<3$ and there are no further solutions in integers.

Thus the only solutions are (1,1), (16,2), and (27,3).

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Zander: Thank you so much and excellent approach. –  gandhi Apr 10 '12 at 4:16

When no one is answering my question. I tried my self and I got the same answer what Zander got. Okay! let me explain.

if d$v^2$ > u, the equation becomes d^(dv^2-u) . u^(dv^2) = $v^u$ => u|v, so u = 1. Similarly we have; d^(dv^2-1) = v. In this case d= 1 gives us v = 1 and x = 1= y as we said earlier. For d > or = 2, there is no solution as d^(dv^2-1) > or = 2^(2v^2-1) > v.

if d$v^2$ < u, the equation becomes u^(dv^2)=d^(u-dv^2) v^u => v|u, so v = 1. Similarly,

$u^d$ = d^(u-d).....(1)

Note that, d = d$v^2$ < u, from (1) we have d < u - d => u > 2d. write u = (p_1)^(t_1)...(p_n)^(t^n) and d = (p_1)^(s_1)...(p_n)^(s^n) and putting in (1) and comparing powers we see;

d(t_i)= (u-d)(s_i) => (s_i) < (t_i) for all i => d|u. write u = kd, where k > or = 3, is an integer from (1), we have; k = d^(k-2). If k = 3, we see d = 3, u = 9, v = 1, x = 27, y = 3. Then we have (27, 3 ) is one of the solution.

If k = 4, we have d= 2, u = 8, v = 1, x = 16 and y = 2. So we have (16, 2) as the other solution. Also, when k = 5 or > 5, there is no solution as d^(k-2)> or = 2^(k-2) > k.

Finally, we have only three solutions. Those are (1, 1), (27, 3) and (16, 2)

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