This is a follow up question to the one posted here. (The method posted here works for all cases where there isn't a coefficient in front of $b$ in the denominator)

Here is the problem: $$\frac{20b}{19b-20}=a$$ $a$ and $b$ have to be positive integers. Prove whether the equation is possible or not possible. By the way, the answer is that it isn't possible.

What I've attempted
My first method was to check if $20b$ are divisible by $1$ or $5$. This didn't work, as you will see. We know $19b-20$ will never be $1$. I tried to check whether $19b - 20$ will be a multiple of $5$, however this approach was wrong, as it may be a multiple of $5$, but that does not mean that it is a factor of $20b$.

My second attempt was playing with the equation algebraically. I got to $20(b+a)=19(ba)$. This shows that $ba$ has to be even, however it does not prove whether the equation is possible or not.

I don't know how to approach these problems that require you to disprove the whether an expression is divisible, especially if the variables are both present in both the denominator and the numerator. Could someone please provide some assistance in proving/disproving this? Thanks for your time.

  • $\begingroup$ You should watch the video "problem 6" or something like that on the numberphile youtube channel. I seem to recall it being similar. It might give you ideas if you don't want an answer right away. $\endgroup$ – stuart stevenson Dec 12 '17 at 16:29
  • $\begingroup$ Here it is: youtube.com/watch?v=Y30VF3cSIYQ. I will watch it, but I don't think the problem I've posted will be that hard, as it is for high school level proofs. $\endgroup$ – WeavingBird1917 Dec 12 '17 at 16:32
  • $\begingroup$ It might be helpful to think of it as $\dfrac 1a+\dfrac 1b=\dfrac {19}{20}$ so the problem is equivalent to whether $\dfrac {19}{20}$ can be expressed as a sum of two unit fractions (Egyptian fraction). $\endgroup$ – hypergeometric Dec 12 '17 at 16:33
  • $\begingroup$ @hypergeometric That is actually what I am trying to prove: whether or not $\dfrac 1x+\dfrac 1y=\dfrac {19}{20}$ is possible. However I thought it would be easier if I could just prove that the equations of the form above was divisible. Would it actually make it harder? $\endgroup$ – WeavingBird1917 Dec 12 '17 at 16:36
  • $\begingroup$ @WeavingBird1917 - per Wolframalpha, $\dfrac {19}{20}=\dfrac 12+\dfrac 13+\dfrac 19+\dfrac 1{180}$, so looks like it's not possible! Alternatively, plot $20(x+y)=19xy$ and it is clear that when $x=2$, $y$ is not an integer, and when $y=2$, $x$ is not an integer. Also $x=\frac {20}{19}>1$ and $y=\frac {20}{19}>1$ are asymptotes. Hence there are no integer solutions. $\endgroup$ – hypergeometric Dec 12 '17 at 16:41

If $b=1$, then we get $a=-20$ which is not a positive integer.

In the following, $b\ge 2$.

We have $$a=\frac{20b}{19b-20}=\frac{19b-20+b+20}{19b-20}=1+\frac{b+20}{19b-20}$$

So, $\frac{b+20}{19b-20}$ has to be a positive integer.

Then, we have to have $$\frac{b+20}{19b-20}\ge 1\implies b+20\ge 19b-20\implies 18b\le 40\implies b\le 2$$

If $b=2$, then we get $a=\frac{20}{9}$ which is not an integer.

  • $\begingroup$ Wow, that's a very nice way of showing it. Is that how you generally approach divisibility problems like this or is it just for this specific case? $\endgroup$ – WeavingBird1917 Dec 12 '17 at 16:47
  • 1
    $\begingroup$ @WeavingBird1917: Having the form $\frac{px+r}{qx+s}$ where $p\lt q$ often works. $\endgroup$ – mathlove Dec 12 '17 at 16:54

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