Can 2z^n(z^n + x^n + y^n) Ever Be a Perfect Square for Prime n Greater Than 2?

[ramsey2879]

Is it true that $$2z^n(z^n + x^n + y^n)$$ can never be a perfect square if n is a prime greater than 2 and x,y,z are prime to each other?

 

[CRGreathouse]

If there is a perfect square of the form $2x^p(x^p+y^p+z^p)$ for p an odd prime, then $2x(x^p+y^p+z^p)$ is also a perfect square, so consider that instead.

Since x, y, and z are coprime, at most one can be odd. If all three were odd then $2x(x^p+y^p+z^p)\equiv2\pmod4$ and so it is not a perfect square. Thus exactly one of x, y, and z is even -- and without loss of generality, we can assume that z is odd.

 

[robert Ihnot]

ramsey2879: Is it true that $$2z^n(z^n + x^n + y^n)$$
can never be a perfect square if n is a prime greater than 2 and x,y,z are prime to each other?

If x^p + y^p = z^p, then $$2z^p(z^p + x^p + y^p)$$
=$$2z^p(2z^p)$$

However, does this imply the converse? Not at all, consider this case of cubes:
x=2, y=3, z=5:

$$(2)(5^3)(2^3+3^3+5^3)=250x 160 = 25(100)(16)=(200)^2$$

Or look at: $$(2)(19^3)(5^3+14^3+19^3) =(2)(19^4)(151+19^2)=(2^{10})(19^4)$$