4

When I graphed the relation $a^{a+2b}=b^{b+2a}$ , it gives a graph similar to $y=x$. However, the question explicitly states that $a \neq b$. So does that mean that no such $N$ exists ?

What happens when the problem is generalized as $N=a^{ma+nb}=b^{mb+na} $ ?

Can anybody help as to what should be done ?

Thanks in advance :) .

  • 1
    What are the restrictions on a,b,n? Are they required to be integers? Are they required to be positive? – quasi Jan 26 '17 at 10:35
  • @quasi The question does not specify anything .... What if they are not integers or are negative.. .??? –  Jan 26 '17 at 10:36
  • 1
    Implicit differentiation shows that the solution set of $x^{x+2y}=y^{y+2x}$ has a singularity at the point $x=y=e^3\approx 20.08$. Did you try to plot it in that region? Near that point Mathematica shows something similar to what I posted here. You can also do implicit differentiation the same way. – Jyrki Lahtonen Jan 26 '17 at 10:37
  • @JyrkiLahtonen O Yeah !!! I did not notice that ... Thanks –  Jan 26 '17 at 10:39
  • @Nirbhay -- where is the question from? – quasi Jan 26 '17 at 10:40
  • @quasi This might prove to be the greatest treasure for you : http://imomath.com/pcpdf/f1/f41.pdf . The question is on Page number 4. :) –  Jan 26 '17 at 10:43
  • @Nirbhay -- thanks for the link. But note -- the problem does say that N is required to be an integer. – quasi Jan 26 '17 at 10:50
  • @JyrkiLahtonen but if we can use negative integers too, couldn't you keep a = 1, b = -2 and get N as 1? – Hiten Jan 26 '17 at 10:51
  • @user237444 I am not sure, but my guess would be that we are not constrained to integers (making number-theory tags kinda misplaced). That also has the side effect that we probably want to assume $x>0,y>0$. For otherwise the powers are not defined unless $a,b$ are integers. Anyway, I'm quite prepared to be wrong about those assumptions. We need the problem source. – Jyrki Lahtonen Jan 26 '17 at 10:55
  • Something like this may help us here. – Jyrki Lahtonen Jan 26 '17 at 10:58
  • 1
    Assuming $a,b$ were intended to be positive, a brute force search yields the following positive integer solution pairs:

    $$(a,b) = (16,32)$$

    $$(a,b) = (16,64)$$

     – quasi Jan 26 '17 at 11:06
  • Nice find @quasi! This is a more complicated curve than the $x^y=y^x$ because the other component is not the graph of a function. – Jyrki Lahtonen Jan 26 '17 at 11:14
  • I don't have a definite proof, but it seems likely to me that (assuming $x>0,y>0$) the minimum of $N$ along that other component curve is achieved at $a=b=e^3$. At least the derivative of $\ln N$ w.r.t. $x=\ln a$ vanishes there. That is bad news in the sense that it would mean that $N$ only has an infimum outside the line $a=b$. After all, the point $a=b=e^3$ is excluded from the reckoning. Also, I don't know if the said derivative has other zeros. I need to go do some real work for a change :-) – Jyrki Lahtonen Jan 26 '17 at 11:48
  • 2
    I can prove that the only positive integer solutions to the diophantine equation $$a^{a+2b}=b^{b+2a}$$ are $(a,b) = (16,32)$ and $(a,b) = (16,64)$. I'll post it as a challenge problem in a separate question. – quasi Jan 26 '17 at 11:48
  • Here is a link to the diophantine version:

    http://math.stackexchange.com/questions/2114905

     – quasi Jan 26 '17 at 12:13
  • @quasi Having a nice time with 'my' problem, aren't you ??? XD –  Jan 26 '17 at 12:29
  • @Nirbhay: Yes, thanks (and I just gave you +1). Although I wasn't able to solve the inequality problem, it did cause me to become curious about the associated diophantine equation. – quasi Jan 26 '17 at 12:36
  • A parametrization of that other component curve seems to be $$a=t^{(t+2)/(t-1)},\quad b=t^{(2t+1)/(t-1)}.$$ Here $t=b/a$, so the two integer points quasi found come from $t=2$ and $t=4$ respectively. Plotting $d(\log N)/dt$ shows that the only zero of that derivative is at $t=1$ corresponding to the case $a=b$ (so on this component $a=b=e^3$). After all, if $t=1+1/u$, then $a=(1+\dfrac1u)^{3u+1}$ that clearly $\to e^3$ when $u\to\infty$. – Jyrki Lahtonen Jan 26 '17 at 13:46
  • So as a summary it may well turn out that quasi's Diophantine version is what was intended? – Jyrki Lahtonen Jan 26 '17 at 13:57
  • I looked at your source. There it is specified that $N$ should be an integer. In that case it is either $2^{80}$ or $\lceil e^{9e^3}\rceil$ depending on whether $a$ and $b$ are required to natural numbers or not. – Jyrki Lahtonen Jan 26 '17 at 21:39
  • Does anyone know the answer ? –  Jan 27 '17 at 14:41

0 Answers0