Equations of the form are called Mordell equations. In this post we’re going to prove that the equation has no integer solutions, using (with one exception) nothing more complicated than congruences.
Theorem: There are no integer solutions to the equation .
Case 1. First, suppose that there is a solution in which x is even. Since , . Looking at the equation modulo 8, then, we have
However, if we square the residues and reduce them modulo 8 we have . So there is no integer y that when squared is congruent to 7 modulo 8. Therefore, there is no solution to when x is even.
Case 2. Now, suppose there is a solution in which x is odd. If so, then we have or .
If then we have . However, if we square the residues and reduce them modulo 4 we get . So there is no integer y that when squared is congruent to 2 modulo 4. Thus there is no solution to when .
Now, suppose . Applying some algebra to the original equation, we have
By assumption, . If has prime factors that are all equivalent to 1 modulo 4, then their product (i.e., ) would be equivalent to 1 modulo 4. Thus has at least one prime factor q that is congruent to 3 modulo 4.
However, if , then . This means that . However, thanks to Euler’s criterion, only odd primes q such that can have a solution to . Since , has no solution. Thus there is no solution to when , either.
Since we’ve covered all cases, there can be no solution in integers to the Mordell equation .