Writing a square as a sum of three non-zero squares in geometric progression 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!
 A: 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.
A: According to  Will Jagy's answer, a parametric solution of $a^2+ ac +c^2=k^2$ is given as follows.
$(a,c)=(u^2-v^2,2uv+v^2)$.
Hence we get $b^2=2vu^3+v^2u^2-2v^3u-v^4$.
Above equation can be transformed to an elliptic curve below where $X=2u/v$ .
$$Y^2 = X^3+X^2-4X-4$$
According to LMFDB, this elliptic curve has rank $0$ and has three integer solutions $(X,Y)=(-2, 0), (-1, 0), (2, 0)$.
Hence the problem has no positive integer solution.
