Diophantine equation x^4 - y^4 = 2 z^2

[TTob]

I need to prove that the equation x^4 - y^4 = 2 z^2 has no positive integer solutions.

I have tried to present this equation in some known from (like x^2+y^2=z^2 with known solutions, or $$x^4 \pm y^4=z^2$$ that has no integer solutions) without success.

Any hints ?

 

[Petek]

I think I have a solution. I'll post a hint and take another look tomorrow. Rewrite

[tex]x^4 - y^4 = 2z^2[/itex]

as

$$[(x + y)^2 + (x - y)^2](x + y)(x - y) = (2z)^2$$.

Then let

$$X = x + y, Y = x - y, Z = 2z$$

so we have

$$XY(X^2 + Y^2) = Z^2$$.

Show that this equation has no non-trivial solutions (several additional steps are still required).

Petek

 

[disregardthat]

I don't know if this works, but looking at different residues modulo 8 seems promising.

 

[CRGreathouse]

I don't know if this works, but looking at different residues modulo 8 seems promising.

Yes, it works.

 

[blue_raver22]

I understand the importance of using evidence and logical reasoning to support claims. In this case, we can approach this problem by using the properties of Diophantine equations and the properties of even and odd numbers.

Firstly, a Diophantine equation is an equation where the solutions are restricted to integers only. In this case, we are looking for positive integer solutions for the equation x^4 - y^4 = 2 z^2. This means that all three variables, x, y, and z, must be positive integers.

Next, let's consider the properties of even and odd numbers. An even number can be written as 2k, where k is any integer. Similarly, an odd number can be written as 2m+1, where m is any integer. This means that the square of an even number will also be even, and the square of an odd number will be odd.

Now, let's look at the equation x^4 - y^4 = 2 z^2. Since both x and y are raised to the fourth power, they will always be even numbers. This means that x^4 - y^4 will always be an even number, and therefore, z^2 must also be an even number. This means that z must also be an even number.

However, this leads to a contradiction because if z is even, then 2 z^2 will be divisible by 4. But x^4 - y^4 is not divisible by 4, as both x and y are even. This means that the equation x^4 - y^4 = 2 z^2 cannot have any positive integer solutions.

In conclusion, using the properties of Diophantine equations and even and odd numbers, we have shown that the equation x^4 - y^4 = 2 z^2 has no positive integer solutions. This can be considered as strong evidence that the equation is not solvable in positive integers.