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?
 A: 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.
A: Here is the solution in Ireland and Rosen (page 270). 
Suppose the equation has a solution. Then $x$ is odd. For otherwise reduction modulo 4 would imply 3 is a square modulo 4.  Write the equation as 
$$y^2+1=(x+2)(x^2-2x+4)=(x+2) ((x-1)^2+3) \ . \ \ \  (*)$$
Now since $(x-1)^2 +3$  is of the form $4n+3$ there is a prime $p$ of the form $4n+3$ dividing it and reduction of $(*)$  modulo $p$  implies that $-1$ is a square modulo $p$  which is a contradiction. 
A: LHS being a perfect square must have "digital  root" $1$,$4$,$7$ or $9$. Cubes have digital root $1$,$8$ or $9$. Hence RHS doesn't have same digital roots as LHS so can't be equal.
