We begin by assuming, for the sake of contradiction, that there exists a non-trivial integer solution (x, y, z) to the equation x^4 + y^4 = z^4. 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.

We begin by assuming, for the sake of contradiction, that there exists a non-trivial integer solution (x, y, z) to the equation x^4 + y^4 = z^4. 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.

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.

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.

Let p = a² and q = b² for some integers a, b. Then p² + q² = (a²)² + (b²)² = a^4 + b^4. Since p² + q² must also be a perfect square, let p² + q² = c². This leads to a^4 + b^4 = c². This is an equation of the same form as the original, x^4 + y^4 = z^4, but with z² instead of z^4. More importantly, z² = m² + n² = (p² + q²)² + (2pq)² = c^4 + (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 a^4 + b^4 = c².

Let p = a² and q = b² for some integers a, b. Then p² + q² = (a²)² + (b²)² = a^4 + b^4. Since p² + q² must also be a perfect square, let p² + q² = c². This leads to a^4 + b^4 = c². This is an equation of the same form as the original, x^4 + y^4 = z^4, but with z² instead of z^4. More importantly, z² = m² + n² = (p² + q²)² + (2pq)² = c^4 + (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 a^4 + b^4 = c².

Prove that the equation x⁴ + y⁴ = z⁴ has no non-trivial integer solutions (special case of Fermat's Last Theorem).

Answer: The equation $x^4 + y^4 = z^4$ has no non-trivial integer solutions.

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 $x^4 + y^4 = z^4$ . A non-trivial solution means $x, y, z \neq 0$ . 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 $(x^2)^2 + (y^2)^2 = (z^2)^2$ . This form shows that $(x^2, y^2, z^2)$ 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 $(x^2, y^2, z^2)$ is a primitive Pythagorean triple (because $x, y, z$ are coprime, so $x^2, y^2, z^2$ must also be coprime), there exist coprime integers $m > n > 0$ of opposite parity such that $x^2 = m^2 - n^2$ , $y^2 = 2mn$ , and $z^2 = m^2 + n^2$ . We can swap $x$ and $y$ if needed to ensure $y^2$ is the even term.

Step 4: Analyze $x^2 = m^2 - n^2$

From $x^2 = m^2 - n^2$ , we can rearrange it to $x^2 + n^2 = m^2$ . 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 $x = p^2 - q^2$ , $n = 2pq$ , and $m = p^2 + q^2$ .

Step 5: Substitute into $y^2 = 2mn$

Now substitute the expressions for $m$ and $n$ into the equation for $y^2$ : $y^2 = 2mn$ . This gives $y^2 = 2(p^2 + q^2)(2pq) = 4pq(p^2 + q^2)$ . Since $y^2$ is a perfect square, $4pq(p^2 + q^2)$ must also be a perfect square. Since $p$ and $q$ are coprime and of opposite parity, $p$ , $q$ , and $p^2 + q^2$ are pairwise coprime. For their product $4pq(p^2 + q^2)$ to be a perfect square, each factor $p$ , $q$ , and $p^2 + q^2$ must individually be a perfect square.

Step 6: Derive a smaller solution

Let $p = a^2$ and $q = b^2$ for some integers $a, b$ . Then $p^2 + q^2 = (a^2)^2 + (b^2)^2 = a^4 + b^4$ . Since $p^2 + q^2$ must also be a perfect square, let $p^2 + q^2 = c^2$ . This leads to $a^4 + b^4 = c^2$ . This is an equation of the same form as the original, $x^4 + y^4 = z^4$ , but with $z^2$ instead of $z^4$ . More importantly, $z^2 = m^2 + n^2 = (p^2 + q^2)^2 + (2pq)^2 = c^4 + (2ab)^4$ . The new solution $(a, b, c)$ has $c^2 = p^2 + q^2 = m < m^2 + n^2 = z^2$ . This means $c < z$ . We have found a smaller positive integer solution $(a, b, c)$ to $a^4 + b^4 = c^2$ .

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 $x^4 + y^4 = z^4$ has no non-trivial integer solutions.

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