探究丢番图方程$x^4+2y^4=z^2$的非平凡解存在性

Great question! This is a classic Diophantine equation, and we can use the method of infinite descent (the same tool that proves $x^4 + y^4 = z^2$ has no non-trivial solutions) to show it has no non-trivial integer solutions either. Let's walk through the reasoning step by step.

First, let's define "non-trivial" as integer solutions where $x, y, z$ are not all zero. We can focus on primitive solutions (where $\gcd(x,y)=1$) since any non-primitive solution can be scaled down to a primitive one.

Step 1: Assume a minimal primitive solution exists #

Suppose there is a smallest set of positive integers $(x,y,z)$ such that $x^4 + 2y^4 = z^2$ and $\gcd(x,y)=1$.

Step 2: Analyze parity #

Looking at the equation modulo 2: $x^4 \equiv z^2 \pmod{2}$. Since fourth powers and squares are either 0 or 1 mod 2, $x$ and $z$ must share the same parity.

• If $x$ is even, $z$ must also be even. Let $x=2x_1$, $z=2z_1$. Substituting gives:
  $$16x_1^4 + 2y^4 = 4z_1^2 \implies 8x_1^4 + y^4 = 2z_1^2$$
  This implies $y^4$ is even, so $y$ is even—but this contradicts $\gcd(x,y)=1$. Thus, $x$ and $z$ must be odd.

Step 3: Factor in $\mathbb{Z}[\sqrt{2}]$ #

The ring $\mathbb{Z}[\sqrt{2}]$ is a unique factorization domain (UFD), which lets us rewrite the equation as:
$$x^4 = z^2 - 2y^4 = (z - y^2\sqrt{2})(z + y^2\sqrt{2})$$
Since $\gcd(z,y)=1$ and $z$ is odd, the two factors are coprime in $\mathbb{Z}[\sqrt{2}]$. In a UFD, if a product of coprime elements is a fourth power, each element must be a fourth power (up to a unit, which we can absorb into our factorization). So we write:
$$z - y^2\sqrt{2} = (a + b\sqrt{2})^4$$
for some integers $a,b$.

Step 4: Expand and equate coefficients #

Expanding the right-hand side:
$$(a + b\sqrt{2})^4 = (a^4 + 12a^2b2 + 4b^4) + (4a^3b + 8ab^3)\sqrt{2}$$
Equating rational and irrational parts gives:
$$z = a^4 + 12a^2b2 + 4b^4$$
$$-y^2 = 4ab(a^2 + 2b^2)$$
Taking absolute values:
$$y^2 = 4|ab|(a^2 + 2b^2)$$

Step 5: Apply infinite descent #

Since $\gcd(x,y)=1$, $\gcd(a,b)=1$ (otherwise, the common divisor would divide both $x$ and $y$, violating primitivity). The factors $|a|$, $|b|$, and $a^2 + 2b^2$ are pairwise coprime (check gcds: since $\gcd(a,b)=1$, any common divisor would have to divide 2, but $x$ is odd so neither $a$ nor $b$ can both be even).

For the product to be a square, each coprime factor must be a square:

• If $a$ is odd: $|a|=m^2$, $|b|=n^2$, $a^2 + 2b^2=p2$, leading to $m^4 + 2n^4 = p^2$—a smaller solution, contradicting our "minimal" assumption.
• If $a$ is even: Let $a=2k$, substituting gives $(y/4)^2=|kb|(2k2 + b^2)$. Again, each factor must be a square, leading to $2s^4 + t^4 = q^2$—another smaller solution, which also contradicts minimality.

Conclusion #

In all cases, we end up with a smaller solution than our initial "minimal" one, which is impossible. Therefore, $x^4 + 2y^4 = z^2$ has no non-trivial integer solutions (only the trivial solution $x=y=z=0$).

内容的提问来源于stack exchange，提问作者STrick