<s>[INST] For what value of m that the equation $y^2 = x^3 + m$ has no integral solutions? For what value of m does equation $y^2 = x^3 + m$ has no integral solutions?
 [/INST] None of the solutions posted look right (I don't think this problem admits a solution by just looking modulo some integer, but possibly I'm wrong). Here is a proof.
First, by looking modulo $8$ one deduces we need $x$ to be odd.
Note that $y^2 + 1^2 = (x+2)(x^2 - 2x + 4)$. As the LHS is a sum of two squares, no primes $3 \pmod{4}$ divide it. This forces $x \equiv 3 \pmod{4}$ as if $x \equiv 1 \pmod{4}$ then $x+2$ obviously has a prime factor $3 \pmod{4}$. But then $x^2 - 2x + 4 \equiv 3 \pmod{4}$, implying  $x^2 - 2x + 4$ has a prime factor $3 \pmod{4}$. But this is a contradiction, thus no $x,y$ can exist to satisfy this equation.
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