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$x^3+7=y^2$ find all solutions in natural numbers

I tried to check a residue of the division on $3,4,5,7,9,12,13$

I want to prove that this equation has not solutions because left part gives some residue and right part other. Actually I don't know is it true . And am I on right way of solving this problem? But I proved this for $x^3+7=y^4$ by checking residues when dividing on $13$ If I'm on right way ,can you say division of what I need to check , to get the result? If not tell me please what to do.

Delta Account
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1 Answers1

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$y^2 = x^3 + k$ is known an Mordell's equation.

You are right that there are no integer solutions. However, I believe that a naive modulo residue check doesn't yield

Here's a suggested approach, which requires you to fill in some gaps

  1. Assume that an integral solution exists. Show that $x$ is odd.
  2. Write $ y^2 + 1 = x^3 + 8 = (x+2)(x^2 - 2x + 4 )$
  3. Hence there is a prime $p$ such that $ p \equiv 3 \pmod{4}$ and $ y^2 + 1 \equiv 0 \pmod{p}$, which is a contradiction.

Note: There are 2 ways of doing 3, based on what is written (and some gaps).

Calvin Lin
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