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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