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

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

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

- (2a)
*x*^{2}= 2*pq* - (2b)
*y*^{2}=*p*^{2}−*q*^{2} - (2c)
*z*=*p*^{2}+*q*^{2}

(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*= 2*ab* - (3b)
*y*=*a*^{2}−*b*^{2} - (3c)
*p*=*a*^{2}+*b*^{2}

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

- (4a)
*x*^{2}= 2*pq*= 2(*a*^{2}+*b*^{2})(2*ab*) = 4(*ab*)(*a*^{2}+*b*^{2})

(5) *ab* and *a*^{2}+*b*^{2} are coprime, because if prime *p* divides *ab* it divides *a* or *b* but not both, so *p* doesn't divide *a*^{2}+*b*^{2}. If the product of coprime integers is a square, then both integers are squares, so *ab* and *a*^{2}+*b*^{2} are both squares. By the same reasoning, *a* and *b* must also be squares.

(6) Since *a*^{2}+*b*^{2} is a square, an integer *P* exists such that

- (6a)
*P*^{2}=*a*^{2}+*b*^{2}

(7) Combining (6a), (3c) and (2c), and noting that *p* < *p*^{2}+*q*^{2} and, since *z*>1, *z* < *z*^{2},

- (7a)
*P*^{2}=*a*^{2}+*b*^{2}=*p*<*p*^{2}+*q*^{2}=*z*<*z*^{2}

(8) We have an infinite descent now, because we started with Pythagorean triple (*x*^{2}, *y*^{2}, 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