Are there positive integers that make the following equation true? $x^2-y^{11}=23$
I do not think it is possible because I have tried many numbers any none of them seem to work. I was able to solve it with an easier case: $x^2 – y^2 = 23$. We can factor to get: $(x+y)(x-y)=23$ Since 23 is prime, its prime factorization is $23 \cdot 1$. We can now solve for $x$ any by doing $(x+y) = 23$ and $(x-y)=1$. $x = 12$, $y = 11$. Although this works for this case, I don't know if it'll work for the original problem.
Best Answer
Suppose a solution exists. First note that $x$ has to be even. Indeed, if it was odd, then $y^{11}=x^2-23$ would be even but not divisible by $4$ (look mod $4$), which is impossible.
It follows that $y$ has to be odd. But then, since $y^{11}$ is congruent to $y$ modulo $4$ for odd $y$, we find that $y\equiv 1\pmod 4$.
Now comes the trick. By adding $y^{11}+2025$ to both sides, we find $$x^2+45^2=y^{11}+2^{11}=(y+2)(y^{10}-2y^9+2^2y^8\pm\dots+2^{10})=(y+2)A.$$ Observe that A is relatively prime to $y+2$ - we have $A=y^{10}-2y^9+2^2y^8\pm\dots+2^{10}\equiv 11(-2)^{10}\pmod{y+2}$, so since $y+2$ is odd, the only common factor could be $11$, which requires $y\equiv -2\pmod{11}$. But going back to the original equation, this would imply $-1$ is a square modulo $11$, which it certainly isn't.
To sum up, we have established that $y+2$ and $A$ are relatively prime numbers whose product is a sum of two squares. However, we can see that both of them are $3\pmod 4$, so both have prime factors which are $3\pmod 4$. One of those factors must not be equal to $3$, call it $p$. By considering the displayed equation modulo $p$, we find $x^2\equiv -45^2\pmod p$, and since $p\neq 3,5$, by taking multiplicative inverse of $45$ modulo $p$ we find $z^2\equiv -1\pmod p$ has an integer solution. However, it is well-known that this is impossible whenever $p\equiv 3\pmod 4$. This gives a contradiction, showing that the equation has no solution.
(Proof inspired by the proof of Theorem 2.1 here.)