Complex Extensions, Two Squares and a Fourth

Two Squares and a Fourth

This section characterizes integer solutions when two squares and a fourth power are combined. This assumes you are familiar with the characterization of the pythagorean triples.

x2 + y4 = z2 (y even)

Apply the coprime lemma, and search for a solution where the variables are pairwise coprime. Thus x and z are odd.

Each term is a square, so characterize the variables as follows.

u > v > 0, u and v coprime, and u and v have different parity:
x ← u2 - v2
y2 ← 2uv
z ← u2 + v2

Since u and v are coprime, the second equation implies u and v are an odd square and twice a square. Conversely, any u and v meeting these criteria produce a solution.

Acceptable values for u and v are: 1, 2, 4, 8, 9, 16, 18, 25, 32, 36, 49, 50, etc. Here are the first four examples, when u and v are drawn from 1, 2, 8, and 9.

32 + 24 = 52
632 + 44 = 652
772 + 64 = 852
172 + 124 = 1452

x2 + y4 = z2 (y odd)

The characterization tells us y2 = u2-v2, or y2+v2 = u2. This makes v even. Characterize these triples as follows.

s > t > 0, s and t coprime, and s and t of different parity:
u ← s2 + t2
v ← 2st
x ← 4st(s2+t2)
y ← s2-t2
z ← s4 + 6s2t2 + t4

The first example, with s=2 and t=1, is 402 + 34 = 412.

x2 + y2 = z4

characterize the triples, whence u2 + v2 = z2. Characterize these triples, and u = s2-t2 and v = 2st. This does not assure u > v, but if v > u, x will come out negative, and that's no big deal; just make it positive.

s > t > 0, s and t coprime, and s and t of different parity:
x ← s4 - 6s2t2 + t4
y ← 4st(s2-t2)
z ← s2 + t2

The first example, with s=2 and t=1, is 72 + 242 = 54.

Set s = 3 and t = 2 to get 1192 + 1202 = 134.