If $x,y,z, a$ are non-zero integers such that
$$\frac{x}{z}=\frac{x^2+y^2}{z^2 + a^2}$$
why can't $x^2+y^2+z^2$ be prime? We assume that $x \ne z$.
This seems to hold but I can't see how to prove it.
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$\begingroup$ 1 + 36 + 4 = 41 which is prime $\endgroup$– DanJan 8, 2014 at 9:58
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2$\begingroup$ @user115998: If there is no restriction on $a$, there would be a lot of counterexamples... $\endgroup$– mathloveJan 8, 2014 at 10:01
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2 Answers
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I'm afraid it can be prime.
If $\left ( x,y,z,a \right )=\left ( 5,15,1,7 \right )$ then $$\frac{x}{z} = \frac{5}{1} = 5$$ $$\frac{x^2+y^2}{z^2+a^2}=\frac{5^2+15^2}{1^2+7^2}=\frac{250}{50}=5$$ and $$x^2+y^2+z^2 = 5^2+15^2+1^2=251$$
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$x^2+y^2=\dfrac{x(z^2+a^2)}{z} \implies z=pq, x=mp,z^2+a^2=nq ,a^2=(n-p^2q)q\implies \\case 1:n=hq \implies x^2+y^2=mhq \implies x^2+y^2+z^2=(mh+p^2)q \\case2: q=q'^2 \implies x^2+y^2+z^2=h'q' $
if $q=1$,then $x=k'z$ or $ z^2+a^2=k''z$, I left these two cases for op.
