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Show that the equation $y^2 = x^3 + 7$ has no integral solutions.

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4  
What have you tried? –  Chris Eagle Oct 27 '12 at 14:43
2  
hint: compute modulo 4. –  Raymond Manzoni Oct 27 '12 at 15:00

3 Answers 3

up vote 5 down vote accepted

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.

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+1: This method was also proposed in Conrad's paper about the more general Mordell equation with a variant. –  Raymond Manzoni Oct 28 '12 at 8:28
    
@RaymondManzoni thanks for the link. It looks very interesting. –  PAD Oct 28 '12 at 10:12

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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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.

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If digital roots work like I think they do, can't both sides have a digital root of 7? –  Mike Oct 27 '12 at 20:10

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