Application of Class Groups

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)(xr), where r^2 = -5. If I is an ideal of R = Z[SQRT(-5)] containing both, then it must contain their sum, 2x, and their difference, 2r. The norm of I must then divide N(2r) = 20 and N(2x) = 4x^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 xr, 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 3k times, where k is an integer. Hence every prime ideal occurs in each of xr and x+r a number of times divisible by 3, and we can find ideals I and J such that (xr) = I^3 and (x+r) = J^3.

Also recall that R has class number 2, or, equivalently, a class group isomorphic to Z/2Z. 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 – 15ab^2 – 5(b^3)r). Equating the coefficients of r, we get (+/-)r = 3(a^2)br – 5(b^3)r so that b*(3a^2 – 5b^3) = (+/-)1. Since a and b are integers, we must have b = (+/-)1.

If b = 1, then 3a^2 – 5 = (+/-)1, which is impossible since 3a^2 can be 0, 3, 12, 27… but not 4 or 6. If b = -1, then 3a^2 + 5 = (+/-)1, which is again impossible since 3a^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 + 5b^2, which is true iff a = (+/-)1 and b = 0.

One Response to Application of Class Groups

  1. […] 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 […]

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