<s>[INST] Existence of integers s.t. $Y^2 = X^3 + n^3 - 4m^2$ Given $m, n \in \mathbb Z$ with $n = -1 (mod 4)$. Let's say that $m$ has no prime divisors that are congruent to $3 (mod 4)$. 
Why are there no integers $X, Y$ such that $Y^2 = X^3 + n^3 -4m^2$ ? 
 [/INST] Assume for the sake of contradiction that a solution $(X,Y)$ exists.  Write $$Y^2+(2m)^2=X^3+n^3=(X+n)(X^2-nX+n^2).$$  We first claim that any prime factor $p$ of $Y^2+(2m)^2$ satisfies $p=2$ or $p\equiv 1\pmod{4}$.  
To show this, let $p\equiv 3\pmod{4}$ be a prime factor of $Y^2+(2m)^2$.  Because $\left(\frac{-1}{p}\right)=-1$, we conclude that $p$ divides both $Y$ and $m$.  But this contradicts the assumption that $m$ has no prime divisor congruent to $3$ modulo $4$.
If $Y^2+(2m)^2$ is even, then $Y$ is even, and therefore $X^3+n^3$ is even and divisible by $4$.  Because $X^2-nX+n^2$ is always odd regardless of the parity of $X$, $4$ divides $X+n$.  Because $n\equiv-1\pmod 4$, we get $X\equiv 1\pmod 4$ so $$X^2-nX+n^2\equiv 3\pmod{4}.$$  Since $X^2-nX+n^2>0$, $X^2-nX+n^2$ has a prime divisor congruent to $3$ mod $4$.  This is a contradiction. 
Assume now that $Y^2+(2m)^2$ is odd.  Then $X$ must be even.  Since $X+n>0$ is odd, we get $X+n\equiv 1\pmod{4}$.  Thus, $X\equiv 2\pmod{4}$, but then
$$X^2-nX+n^2\equiv 3\pmod{4}.$$
This is another contradiction.
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