## No Integer Solutions to a Mordell Equation

Equations of the form $x^3 = y^2 + k$ are called Mordell equations.  In this post we’re going to prove that the equation $x^3 = y^2 -7$ has no integer solutions, using (with one exception) nothing more complicated than congruences.

Theorem: There are no integer solutions to the equation $x^3 = y^2 -7$.

Proof.

Case 1.  First, suppose that there is a solution in which x is even.  Since $2|x$, $8|x^3$.  Looking at the equation modulo 8, then, we have

$\displaystyle 0 \equiv y^2 - 7 \pmod{8} \implies y^2 \equiv 7 \pmod{8}$.

However, if we square the residues $\{0, 1, 2, 3, 4, 5, 6, 7\}$ and reduce them modulo 8 we have $\{0, 1, 4, 1, 0, 1, 4, 1\}$.  So there is no integer y that when squared is congruent to 7 modulo 8.  Therefore, there is no solution to $x^3 = y^2 -7$ when is even.

Case 2.  Now, suppose there is a solution in which x is odd.  If so, then we have $x \equiv 3 \pmod{4}$ or $x \equiv 1 \pmod{4}$.

If $x \equiv 3 \pmod{4}$ then we have $y^2 \equiv 27 + 7 \equiv 34 \equiv 2 \pmod{4}$.  However, if we square the residues $\{0, 1, 2, 3\}$ and reduce them modulo 4 we get $\{0, 1, 0, 1\}$.  So there is no integer y that when squared is congruent to 2 modulo 4.  Thus there is no solution to $x^3 = y^2 - 7$ when $x \equiv 3 \pmod{4}$.

Now, suppose $x \equiv 1 \pmod{4}$.  Applying some algebra to the original equation, we have

$\displaystyle y^2 = x^3 + 7 \implies y^2 + 1 = x^3 + 8 = (x+2)(x^2 - 2x + 4)$.

By assumption, $x + 2 \equiv 3 \pmod{4}$.  If $x + 2$ has prime factors that are all equivalent to 1 modulo 4, then their product (i.e., $x+2$) would be equivalent to 1 modulo 4.  Thus $x + 2$ has at least one prime factor q that is congruent to 3 modulo 4.

However, if $q|x+2$, then $q|y^2+1$.   This means that $y^2 \equiv -1 \pmod{q}$.  However, thanks to Euler’s criterion, only odd primes q such that $q \equiv 1 \pmod{4}$ can have a solution to $y^2 \equiv -1 \pmod{q}$.  Since $q \equiv 3 \pmod{4}$, $y^2 \equiv -1 \pmod{q}$ has no solution.  Thus there is no solution to $x^3 = y^2 - 7$ when $x \equiv 1 \pmod{4}$, either.

Since we’ve covered all cases, there can be no solution in integers to the Mordell equation $x^3 = y^2 + k$.

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