Do any whole number solutions exist for $a^2 + b^3 = 2^{2023}$?
And if so, would it be possible to give an example of one?
This problem is meant to be solved with modular-arithmetic and I could not even get started. Guessing that you must find the right modulus such that no squares or cubes can be added to reach the mod of $2^{2023}$.
For example all squares in mod $8$ are $0$, $1$ or $4$. And all cubes have in mod $13$ class $0$, $\pm1$, $\pm 5$.
Any help appreciated!
The question is
One potential solution would be if $a^2 = b^3 = 2^{2022}$. Note that this has obvious solutions $a = 2^{2022/2} = 2^{1011}$ and $b = 2^{2022/3} = 2^{674}.$ There may be other non-obvious solutions.
This is an example of an elliptic curve. The base case in standard Weierstrass form is
$$ y^2 = x^3 + 2 \tag1 $$
which is the curve with Cremona label 1728a1. This curve has the obvious points $(x,y) = (-1,\pm 1).$ Each is a generator for the group of rational points. All other rational points are not integral.
In your case, you want something like
$$ y^2 = x^3 + 2^{1+6k} \tag2 $$
or equivalently,
$$ (y/8^k)^2 = (x/4^k)^3 + 2. \tag3 $$
The problem now becomes to find a solution $(x,y)$ of equation $(1)$ where the denominator of $x$ is $4^k$ which seems to be impossible if $k>0$.