Prove that the Diophantine equation x4y4z2 has no solutions

Prove that the Diophantine equation x^{4}-y^{4}=z^{2} has no solutions in nonzero integers x, y, z.

Solution

Suppose z2=y4x4z2=y4x4 with xyz0 for the smallest possible value of z2. First we rewrite the equation as y4=x4+z2so that {z,x2,y2} is a Pythagorean triple. It must be primitive, since if some prime p divides gcd(x2,y2) then p|y2 implies p|y which gives p4|y4. Similarly, p4|x4, so p4|z2. This implies p2|z, so that (y/p)4(x/p)4=(z/p2)2 is a smaller solution.

The Pythagorean triple z,x2,y2 is primitive and there are two cases:

If x is even, then for some m>n, (m,n)=1(, and mn(mod2)we have

z=m2n2,x2=2mn,y2=m2+n2.

ince m,nm,n have opposite parity, we can let oo denote the odd number and ee the even number among {m,n}{m,n}. The primitive Pythagorean triple {n,m,y}{n,m,y} gives

o=t2s2,e=2st,y=t2+s2,

or some t>st>s, (t,s)=1(t,s)=1, and ts(mod2)ts(mod2). The formula for x2x2 now gives

(x/2)2=ts(t2s2)

which expresses the product of three relatively prime numbers as a square. That means all three of them are squares: s=u2s=u2, t=v2t=v2, and t2s2=w2t2s2=w2. In other words, v4u4=w2v4u4=w2 is another solution to our equation, and it is smaller, since v4<t2+s2=yy4v4<t2+s2=yy4.

If xx is odd, then for some m>nm>n, (m,n)=1(m,n)=1, and mn(mod2)mn(mod2) we have

x2=m2n2,z=2mn,y2=m2+n2.x2=m2n2,z=2mn,y2=m2+n2.

In this case m4n4=(xy)2m4n4=(xy)2 is a smaller solution, since m4<(m2+n2)2=y4m4<(m2+n2)2=y4.