Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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

2 Answers 2

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).

share|improve this answer
    
Zander: Thank you so much and excellent approach. –  gandhi Apr 10 '12 at 4:16
up vote 0 down vote accepted

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)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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