Recall that the proof that there are no integer solutions to the equation *x*^2 + 2 = *y*^3 except *x* = (+/-)5, *y* = 3 is based on unique factorization in the ring **Z**[SQRT(-2)]. Now I’m going to similarly deal with the equation *x*^2 + 5 = *y*^3, where factorization is in the ring **Z**[SQRT(-5)], which has no unique factorization.

First, suppose that the equation has an integer solution. If *x* is odd, then *x*^2 = 1 mod 4, so *y*^3 = 2 mod 4, which means 2 divides *y*^3 but 4 doesn’t. This is clearly a contradiction, so *x* is even. If *x* is divisible by 5, then *y*^3 = 5 mod 25, which means 5 divides *y*^3 but 25 doesn’t. This is again a contradiction, so *x* is not divisible by 5.

Now, let us factor *x*^2 + 5 as (*x*+*r*)(*x*–*r*), where *r*^2 = -5. If I is an ideal of R = **Z**[SQRT(-5)] containing both, then it must contain their sum, 2*x*, and their difference, 2*r*. The norm of I must then divide N(2*r*) = 20 and N(2*x*) = 4*x*^2. Since *x* is not divisible by 5, N(I) must divide 4. But I also contains *x*^2 + 5, which is an odd integer; therefore, N(I) is itself odd. So N(I) = 1, and I = R.

Since no proper ideal of R contains both *x*+*r* and *x*–*r*, the unique factorizations of the two elements into ideals contain no prime ideals in common.

The unique factorization of *y*^3 into prime ideals clearly contains every ideal 3*k* times, where *k* is an integer. Hence every prime ideal occurs in each of *x*–*r* and *x*+*r* a number of times divisible by 3, and we can find ideals I and J such that (*x*–*r*) = I^3 and (*x*+*r*) = J^3.

Also recall that R has class number 2, or, equivalently, a class group isomorphic to **Z**/2**Z**. In particular, for every ideal I, the ideal I^3 = (I^2)*I is in the same ideal class as I. Since I^3 and J^3 are principal, so are I and J.

In other words, (*x*+*r*) = (*a*+*br*)^3, so that *x*+*r* = (+/-)(*a*+*br*)^3 = (+/-)(*a*^3 + 3(*a*^2)*br* – 15*ab*^2 – 5(*b*^3)*r*). Equating the coefficients of *r*, we get (+/-)*r* = 3(*a*^2)*br* – 5(*b*^3)*r* so that *b**(3*a*^2 – 5*b*^3) = (+/-)1. Since *a* and *b* are integers, we must have *b* = (+/-)1.

If *b* = 1, then 3*a*^2 – 5 = (+/-)1, which is impossible since 3*a*^2 can be 0, 3, 12, 27… but not 4 or 6. If *b* = -1, then 3*a*^2 + 5 = (+/-)1, which is again impossible since 3*a*^2 is never -4 or -6.

In other words, every possible integer solution of *x*^2 + 5 = *y*^3 leads to a contradiction, so the equation has no integer solutions.

Note that the step from (*x*+*r*) = (*a*+*br*)^3 to *x*+*r* = (+/-)(*a*+*br*)^3 relies on the fact that R has only two units, 1 and -1. This is in fact true for every ring of integers of **Q**(SQRT(*m*)) where *m* < -3 or *m* = -2. For a proof when *m* = 5, consider the norm: *a*+*br* is a unit iff 1 = N(*a*+*br*) = *a*^2 + 5*b*^2, which is true iff *a* = (+/-)1 and *b* = 0.

[…] field generated by all nth roots of unity. In that case, a trick similar to what I did with the equation x^2 + 5 = y^3 can […]