This problem is on Page 194 of “Number Theory and Its History”, by Oystein Ore. I can find infinitely many solutions by letting $x=1$ so that $z^2-5y^2=1$ which has infinitely many solutions. So my question is, are there solutions with x not equal to one, and how can they be found?
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nonuser
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wayne thompson
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4Note for each $(y_1,z_1)$ integer solutions to $z^2 - 5(y^2) = 1$, then $x^6(z^2) - 5(x^6)(y^2) = x^6 ; \to ; (x^{3}z)^2 - 5(x^{3}y)^2 = x^6$, so $(x_1,x_1^3(y_1),x_1^3(z_1))$ is a solution to $z^2 - 5y^2 = x^6$ for all integers $x_1$. – John Omielan Jul 10 '22 at 05:30
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The equation is not difficult to solve. It's just that the formula will be cumbersome. Before us is the Pythagorean triple. $$(x^3)^2+5y^2=z^2$$ We will write the parametrization as. $$x^3=X^2-5Y^2$$ $$y=2XY$$ $$z=X^2+5Y^2$$
It remains to find the parametrization of this equation. It must be solved. $x^3=X^2-5Y^2$
The solution to this equation is there. Given prime $p$, find solutions to $x^2 + p y^2 = z^3$
The formula turns out to be cumbersome. After substituting one formula into another, you will get parameterization. You'll have to do it yourself.
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If $x_0,y_0$ and $z_0$ is a solution to the equation then we see that $$x=x_0t,\;\;y=y_0t^3,\;\;z=z_0t^3$$ is also a solution for arbitrary integer $t$. So all you need is to find one specific solution $x_0,y_0,z_0$ which is not hard to find.
$x_0= 1$, $y_0=4$ and $z_0= 9$ and thus $x=t$, $y=4t^3$ and $z=9t^3$ which means that there is a solution to the equation for an arbitrary $x$.
nonuser
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