Positive integer solutions of $x^2+21y^2=z^4 $

Question

Can one find all positive integer solutions of $$x^2+21y^2=z^4 ?$$

I am not sure if this is possible. I just saw this problem and this problem came to my mind.

Answer 1

I seem to keep making dumb mistakes; both comments I made on this problem (now deleted) are wrong. To make up for it, here is a complete solution.

Note that, given an integer solution to $x^2+21 y^2 = z^4$, we get a rational solution to $X^2 + 21 Y^2 =1$, by $(X,Y) = (x/z^2, y/z^2)$. Conversely, given a rational $(X,Y)$ obeying $X^2 + 21 Y^2=1$, we can always get back to an integer solution $(Xz^2, Yz^2, z)$ by taking appropriate $z$. (And, by taking multiples of that $z$, we get infinitely many solutions.) So I'll concentrate on finding rationals $(X,Y)$ obeying this conic equation.

Once you have one rational point on a conic, there is a standard method for parameterizing the others. In our example, the one point will be $(-1,0)$. Consider a line through this point with rational slope $\mu$. Explicitly, the equation of this line is $Y=\mu(X+1)$. One of the intersections of this line with the conic is $(-1,0)$. The other one must also be rational.

Explicitly, that other point is found by solving $X^2 + 21 \mu^2 (X+1)^2 =1$ or $21 \mu^2 (X+1)^2 = (1+X)(1-X)$. Dividing out the solution $X=-1$, which is the point we already know, we get $21 \mu^2 (X+1) = 1-X$ or $$X = \frac{1-21 \mu^2}{1+21 \mu^2}.$$ Using $Y=\mu(X+1)$, we get $$Y = \frac{2 \mu}{1+21 \mu^2}$$

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It might be worth saying a little bit about how to unpack this back to integers. Let $\mu=p/q$, for some relatively prime $(p,q)$. Our solutions are of the form $$(z^2 \frac{q^2 - 21 p^2}{q^2+21 p^2}, z^2 \frac{2 pq}{q^2 + 21 p^2}, z)$$ for any $z$ for which the first two terms are integers.

One should probably be able to simplify this a little more: For example, I'm pretty sure that, up to factors of $2$, $3$ and $7$, the least common denominator of $X$ and $Y$ is $p^2 + 21 q^2$. I'll leave for that for those who are interested.