I saw this problem in the Problem-Solving through Problems book by Larson (# 3.3.25b).
I got to here:
$$x \log(x) = y\log\left(1+ \frac yx\right)$$
But I can't seem to find a way to reduce this further.
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$(2,2)$, $(9,18)$. – Oleg567 Aug 23 '13 at 16:01
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Thanks! But how did you prove those numbers? I don't want a complete solution but just a thought-process. – user473973 Aug 23 '13 at 16:02
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I also trial-and-error'd the solution (0,1). – user473973 Aug 23 '13 at 16:03
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Can you change variables, and rearrange it with one variable on the left of the equals sign, and the other variable on the right? – Empy2 Aug 23 '13 at 16:07
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That's what I'm trying to do, but cannot see a way to do it with logarithms. I can get it with y = ... but the ... has one instance of y which I cannot get rid of. – user473973 Aug 23 '13 at 16:08
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How about (1,-2)? – Empy2 Aug 23 '13 at 16:56
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It has to be positive rational solutions. – user473973 Aug 23 '13 at 17:12
1 Answers
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Here's a partial solution. The first part is for general $x,y$ and is incomplete. The second part is a complete solution for integer $x,y$.
Part 1: Set $z=x+y$. Then you may rearrange as $$x^zz^x=z^z$$
Now, these are fractions, so set $x=\frac{a}{b}$, $z=\frac{c}{d}$. Assume without loss that $a,b$ are relatively prime, and so are $c,d$. We get $$\left(\frac{a}{b}\right)^z\left(\frac{c}{d}\right)^x=\left(\frac{c}{d}\right)^z$$ or $$a^z c^xd^z=c^zb^zd^x$$ To clear denominators raise both sides to the $bd$ power, to get $$a^{bc}c^{ad}d^{bc}=c^{bc}b^{bc}d^{ad}$$ Becase $z>x$ we have $bc>ad$. Set $k=bc-ad>0$ for convenience. We divide and get $$a^{bc}d^{k}=b^{bc}c^{k}$$ Finally, we have an expression with positive integers. Because $a,b$ are relatively prime, $a^{bc}|c^{k}$ and $b^{bc}|d^k$. Because $c,d$ are relatively prime, $c^{k}|a^{bc}$ and $d^k|b^{bc}$. Hence $a^{bc}=c^k$ and $b^{bc}=d^k$.
Now, rearrange $a^{bc}=c^{bc-ad}$ to get $c^{ad}=(c/a)^{bc}$. Since $c$ is an integer, $a|c$. Similarly, rearrange $b^{bc}=d^{bc-ad}$ to get $d^{ad}=(d/b)^{bc}$, to conclude $b|d$. These are helpful, but it doesn't give explicitly all rational solutions.
Part 2: If we assume integer solutions (not just rational), then $c=d=1$. Because $a|c$ we may write $c=aq$. Now $a^{bc}=c^k$ becomes $$a^{aq}=(aq)^{aq-a}$$ If $a=1$, then $q=c=1$, which is impossible since $y>x$. Otherwise we take logs, getting $$aq \ln a = a(q-1) (\ln a + \ln q)$$ This rearranges to $$\frac{q}{q-1}\ln a=\ln a + \ln q$$ Divide by $\ln a$ (since $a>1$), subtract 1, and add fractions to get $\frac{1}{q-1}=\frac{\ln q}{\ln a}$. Cross-multiply and exponentiate to get the final solution $$a=q^{q-1}$$ This works for any natural $q$; we take $a=q^{q-1}, c=aq=q^q, b=c-a=a(q-1)$. In particular, this finds Oleg's solutions (from $q=2,3$), but also $(64,192)$ (corresponding to $q=4$), etc.
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