$\frac{ (359\cdot (109+215\cdot x)-1)}{10^x}=y$

Question

Consider the diophantine equation:

$\frac{ (359\cdot (109+215\cdot x)-1)}{10^x}=y$, for x,y positive.

The only solution I found is $x=2$, $y=1935$.

Can it be proven that if there is a solution there are infinitely many other solutions?

The left hand side is strictly positive, strictly decreasing and tends to $0$. That means it is strictly smaller than $1$ for sufficiently large $x$. So there can be only finitely many solutions, and a bit of algebra will give you a (small?) upper bound on $x$ in this way, leaving just some cases to check. — Servaes, Apr 10 '21 at 22:20

score 0 · Answer 1 · answered Apr 10 '21 at 10:43

For $x \in \Bbb{Z^+}, x \geq 7$, you have that
$$10^{x-6} \geq (0.2) x.\tag1 $$

Equation (1) above is immediate in the base case of $x = 7$.

Inductively, as $x \to x+1$, the LHS of equation (1) above is multipled by $(10)$ and the right hand side is multiplied by $\frac{x+1}{x}.$

This means that when $x \geq 7$,

$$[10^x y] + 1 > 10^x > (0.2) [(10)^6]x > 39131 + 77185x = 359(109) + 359(215x).$$

Therefore, no solution is possible for $x \geq 7$.

score -1 · Answer 2 · answered Apr 10 '21 at 10:06

$39130+359\cdot215x=39130+359\cdot43\cdot5x$

$359\cdot43\cdot5x$ must end in $70, 870, 10\,870, 60\,870,160\,870...$

If we look at the last digits, $9\times3=27$ so we see $7\cdot5x$ must end in 70. But then we see the next multiple of $35$ that ends in $70$ is $35\times22$ but by then $10^x$ will be huge and we can say this will likely not have any further positive integer solutions as $10^x$ is exponential while $359\cdot43\cdot5x$ is only linear.

Not a very neat solution, just my initial thoughts.

Also, I have no idea how wolfram alpha has negative solutions for $y$. How can $y$ be negative and $x$ positive? Thoroughly confused by that.

$\frac{ (359\cdot (109+215\cdot x)-1)}{10^x}=y$

2 Answers2