Find integers $x$ and $y$ such that $8^x-9^y=431$ Find integers $x$ and $y$ such that $8^x-9^y=431$
My working:
By taking mod 9 and 16, I got $x$ odd and $y$ even.
Also $8^x>431\implies x\ge 3$
For $x=3$ I got $y=2$

$$ 8^{y+d} - 9^y = 431 $$
$$ 8^d = \frac{431 + 9^y}{8^y} $$
$$ 8^d = (\frac{9}{8})^y+\frac{431}{8^y} $$
$$ 8^d \sim (\frac{9}{8})^y $$
$$ 3d\cdot ln(2) \sim y \cdot(2ln(3)-3ln(2)) $$
$$ d \sim y \cdot(\frac{2\cdot ln(3)-3\cdot ln(2)}{3\cdot ln(2)}) $$
$$ d \sim y \cdot 0.0566416671474374543024926292985... $$
$$ \frac{3}{53} = 0,056603773... $$
First better aproximation, for $d=3$ and $y=53$:
$$ 8^{56} -9^{53} = -1,56579...\cdot10^{48} $$
Etc...
This problem is related to the approximation of $log_2 (3)$ to a rational number.
And as the different rational ones approach.
