MHB Prove that x²+y²+z² isn't prime

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The discussion centers on proving that the expression x² + y² + z² cannot be prime under the condition that x, y, and z are nonzero integers with x not equal to z, and the ratio x/z equals (x² + y²)/(y² + z²). Participants are tasked with providing a mathematical proof to support this claim. The conversation emphasizes the need for a rigorous approach to demonstrate that the sum of the squares of these integers cannot yield a prime number. Overall, the goal is to establish a clear mathematical argument against the primality of the expression.
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Let $x,\,y,\,z$ be nonzero integers, $x\ne z$ such that $\dfrac{x}{z}=\dfrac{x^2+y^2}{y^2+z^2}$.

Prove that $x^2+y^2+z^2$ cannot be a prime.
 
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anemone said:
Let $x,\,y,\,z$ be nonzero integers, $x\ne z$ such that $\dfrac{x}{z}=\dfrac{x^2+y^2}{y^2+z^2}$.

Prove that $x^2+y^2+z^2$ cannot be a prime.

Hello.

\dfrac{x}{z}-1=\dfrac{x^2+y^2}{y^2+z^2}-1

\dfrac{x-z}{z}=\dfrac{x^2-z^2}{y^2+z^2}

\dfrac{1}{z}=\dfrac{x+z}{y^2+z^2}

y^2+z^2=xz+z^2 \rightarrow{}y^2=xz1º) Let \ d \in{\mathbb{Z}}/ \ d|x \ and \ d|z \rightarrow{ } d|(x^2+y^2+z^2)2º) For \ x,z \ coprime \rightarrow{}y^2=a^2b^2/ a,b \in{\mathbb{Z}}/ \ x=a^2 \ and \ z=b^2

x^2+y^2+z^2=a^4+a^2b^2+b^4=(a^2+ab+b^2)(a^2-ab+b^2)

Regards.
 
Good job, mente oscura and thanks for participating!:)
 

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