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!:)
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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