Prove that the equation x⁴ + y⁴ = z⁴ has no non-trivial integer solutions (special case of Fermat's Last Theorem).
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Step-by-Step Solution
Step 1: Assume a solution exists
We begin by assuming, for the sake of contradiction, that there exists a non-trivial integer solution (x, y, z) to the equation x4+y4=z4. A non-trivial solution means x,y,z=0. We can also assume x, y, z are positive integers and coprime, as any common factor can be divided out, and if any variable is zero, the problem reduces to a trivial case.
Step 2: Rewrite the equation
We can rewrite the given equation as (x2)2+(y2)2=(z2)2. This form shows that (x2,y2,z2) would constitute a Pythagorean triple. This is a crucial step as it allows us to use properties of Pythagorean triples.
Step 3: Apply properties of Pythagorean triples
Since (x2,y2,z2) is a primitive Pythagorean triple (because x, y, z are coprime, so x2,y2,z2 must also be coprime), there exist coprime integers m > n > 0 of opposite parity such that x2=m2−n2, y2=2mn, and z2=m2+n2. We can swap x and y if needed to ensure y2 is the even term.
Step 4: Analyze x2=m2−n2
From x2=m2−n2, we can rearrange it to x2+n2=m2. This implies that (x, n, m) is also a primitive Pythagorean triple. Therefore, there exist coprime integers p > q > 0 of opposite parity such that x=p2−q2, n=2pq, and m=p2+q2.
Step 5: Substitute into y2=2mn
Now substitute the expressions for m and n into the equation for y2: y2=2mn. This gives y2=2(p2+q2)(2pq)=4pq(p2+q2). Since y2 is a perfect square, 4pq(p2+q2) must also be a perfect square. Since p and q are coprime and of opposite parity, p, q, and p2+q2 are pairwise coprime. For their product 4pq(p2+q2) to be a perfect square, each factor p, q, and p2+q2 must individually be a perfect square.
Step 6: Derive a smaller solution
Let p=a2 and q=b2 for some integers a, b. Then p2+q2=(a2)2+(b2)2=a4+b4. Since p2+q2 must also be a perfect square, let p2+q2=c2. This leads to a4+b4=c2. This is an equation of the same form as the original, x4+y4=z4, but with z2 instead of z4. More importantly, z2=m2+n2=(p2+q2)2+(2pq)2=c4+(2ab)4. The new solution (a, b, c) has c2=p2+q2=m<m2+n2=z2. This means c < z. We have found a smaller positive integer solution (a, b, c) to a4+b4=c2.
Step 7: Conclusion by infinite descent
We started with a positive integer solution (x, y, z) and derived a new positive integer solution (a, b, c) where c < z. We can repeat this process indefinitely, generating an infinitely decreasing sequence of positive integers z > c > c' > .... This is impossible, as there is a smallest positive integer. Therefore, our initial assumption that a non-trivial integer solution exists must be false. This proves that the equation x4+y4=z4 has no non-trivial integer solutions.