## FANDOM

49 Pages

This article is one of a series of articles on Fermat's Last Theorem. Here, we will show that $x^4+y^4=z^2$ has no solutions when $xyz\ne 0$. (The requirement that xyz be nonzero is equivalent to x>0, y>0, and z>1, since z2 is the sum of two numbers, both at least 1.)

(1) We assume for the purpose of descent that a solution to x4+y4=z2 exists with x2, y2, and z coprime, and x2 even. (If prime p divided all three, then a smaller solution $\left(\frac{x}{p}\right)^4+\left(\frac{y}{p}\right)^4=\left(\frac{z}{p}\right)^2$ would exist. If x2 were odd, then we could interchange x and y WLOG.)

(2) By our assumption, there is a primitive Pythagorean triple, (x2, y2, z). From the principles of Pythagorean triples, there exist coprime opposite-parity integers p and q (with p>q) such that

(2a)   x2 = 2pq
(2b)   y2 = p2q2
(2c)   z = p2 + q2

(3) Equation 2b gives us another primitive Pythagorean triple, (q, y, p), and since y is odd, and one of the first two elements of a Pythagorean triple must be even, it follows that q is even, so there exist coprime integers a and b (with a>b) such that

(3a)   q = 2ab
(3b)   y = a2b2
(3c)   p = a2 + b2

(4) Combining equations (2a), (3a), and (3c),

(4a)   x2 = 2pq = 2(a2+b2)(2ab) = 4(ab)(a2+b2)

(5) ab and a2+b2 are coprime, because if prime p divides ab it divides a or b but not both, so p doesn't divide a2+b2. If the product of coprime integers is a square, then both integers are squares, so ab and a2+b2 are both squares. By the same reasoning, a and b must also be squares.

(6) Since a2+b2 is a square, an integer P exists such that

(6a)   P2=a2+b2

(7) Combining (6a), (3c) and (2c), and noting that p < p2+q2 and, since z>1, z < z2,

(7a)   P2 = a2+b2 = p < p2+q2 = z < z2

(8) We have an infinite descent now, because we started with Pythagorean triple (x2, y2, z), with squares as the two smaller elements, and from this we derived the existence of a smaller Pythagorean triple (a, b, P) with squares as the smaller two elements.

QED