Let $k$ be a given positive integer. I want to solve the following system of Diophantine equations: $$\begin{cases} a^2 + b^2 + c^2 = k^2 \\ b^2 = ac \end{cases}$$ where $a, b, c \in \mathbb{N}$ are non-zero.
There is an OEIS sequence for the numbers which are sums of three non-zero squares, but I don't know of any general expression (some squares, such as 25, are not there).
I know, by substituting the second equation into the first, that $k = m^2 + n^2 + mn$, $a = m^2 - n^2$ and $c = 2mn + n^2$, where $m, n$ are coprime integers, but I don't really know how to use this, since plugging this back into the second equation yields a very complicated looking one (namely, when is the product a square?)
I'm not used to solving Diophantine equations (much less systems of them), but I have tried direct computation and seriously believe that there is no solution.
Are there any hints or techniques for solving this type of problem?
Thanks in advance!
we may demand $ \gcd(a,b,c) = 1$ by dividing through by any common factor. This works because of homogeneity.
Next, $ac=b^2$ implies that $$ a = x^2, \; \; b = xy, \; \; c = y^2 $$
At this point you have $$ x^4 + x^2 y^2 + y^4 = k^2 $$ with $\gcd(x,y) = 1.$ This is the result labelled (7') on page 19 of Mordell, Diophantine Equations. There are only trivial solutions. The proof is about a page, on pages 19 and 20. Worth looking up and going through it in detail. It is just taking the "smallest" positive solution and showing a contradiction. No elliptic curves.