Problem
Solve in positive integers the equation: $49^x + 7^{x-2} \cdot 3^{y+1} - 5 \cdot 9^y = 2023$.
- This problem is from the algebra round of a local high school math competition that has ended.
My Work
Goal: Simplify?!
$7^{2x} + 7^{x-2} \cdot 3^{y+1} - 5 \cdot 3^{2y} = 2023$.
$7^{x-2} \cdot (7^{2x-(x-2)} + 3^{y+1}) - 5 \cdot 3^{2y} = 2023$.
$7^{x-2} \cdot (7^{x+2} + 3^{y+1}) - 5 \cdot 3^{2y} = 7 \cdot 17^2$.
Looks kind of messy... Maybe substitution will help???
Lets say $a=7^x$ and $b=3^y$, we have:
$a^2 + \frac{3ab}{49} - 5b^2 = 7 \cdot 17^2$
I see that $3ab$ has to be divisible by $49$, because all of these terms have to be integers, but the equation still looks kind of messy... Could I get some help on how to continue? Thank you so much!
$$49^x + 7^{x-2} \cdot 3^{y+1} - 5 \cdot 9^y = 2023$$
Since this problem is asking for positive integer solutions, and note $7|2023$. Assume $x\ge 3,$ then $7|49^x, 7|7^{x-2}, $ but $7\nmid5\cdot9^y$, contradiction. Therefore, $x<3$, then we just examine $x=1, x=2$.
For $x=1$, the LHS is not integer, hence doesn't work.
For $x=2 $, we get
$$(5\cdot 3^y+42)(3^y-9)=0$$ Therefore, $$3^y=9\Rightarrow y=2$$