Multivariable Equation: $4ab=5(a+b)$

Question

Find all Natural roots of the following multi-variable equation : $4ab=5(a+b)$

I have tried many handy candidate solutions and it seems there is no SOLUTION! Indeed we should show that it has no root set in Natural numbers.

Answer 1

$$ 4ab-5a-5b=0 $$ $$ 16ab - 20a - 20 b = 0 $$ $$ 16ab - 20 a - 20 b + 25 = 25 $$ $$ (4a-5)(4b-5) = 25 $$ three positive divisors , the possible values $(4a-5,4b-5)$ are $$ (-1,-25) \Longrightarrow a=1, b=-5$$ $$ (-5,-5) \Longrightarrow a=0, b=0$$ $$ (-25,-1) \Longrightarrow a=-5, b=1$$ $$ (1,25) \Longrightarrow a=3/2, b=15/2$$ $$ (5,5) \Longrightarrow a=5/2, b=5/2$$ $$ (25,1) \Longrightarrow a=15/2, b=3/2$$ so the original $a,b$ are integers just the first three times.

Answer 2

Well, $a=0=b$ works, but let's exclude them. Rearrange to get $$\frac45=\frac{a+b}{ab}=\frac 1a + \frac 1b$$

If both a and b are greater than 2 than the right hand is less than or equal to $\frac 23$ so that's out. Hence one of them, let's say a, is 2 (clearly, neither can be 1). But it is easy to see that $a=2$ does not lead to a solution in the natural numbers so there are none.

Answer 3

We have $\dfrac{5b}{4b-5}=a$ which must be an integer

$\implies(4b-5)$ must divide $5b$

Clearly, $b=0$ is a solution

If integer $d\ne0$ divides both $4b-5,5b;d$ must divide $4(5b)-5(4b-5)=25$

$\implies d=\pm1,\pm5,\pm25$

So, $4b-5$ must be one of $\{\pm1,\pm5,\pm25\}$