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!
Best Answer
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.