Upon integers that sum up to perfect squares

Question

I have some issues to find three positive integers that sum up to a perfect square... The complete problem wants you to find a triplet of integers $a,b,c>0$ such that $a+b$, $b+c$, $c+a$ are perfect squares; then adds that the same $a,b,c$ previously found with that property have to be such that $a+b+c$ is a perfect square too.
It's kinda easy to find a solution to the first question since you can arbitrarily choose some integers $k,m,n>0$ and then build $a,b,c$ in terms of the former triplet imposing the system $$\begin{cases}a+b=k^2\\b+c=m^2\\c+a=n^2\end{cases}$$ that leads to $$\begin{cases}a=\frac{k^2+n^2-m^2}2\\b=\frac{m^2+k^2-n^2}2\\c=\frac{n^2+m^2-k^2}2\end{cases}$$ Now, if we say $a+b+c=\lambda^2$ for an integer $\lambda>0$, it follows that $2\lambda^2=k^2+m^2+n^2$ and here comes my problem: excluding banal solutions involving pythagorean triplets (that also make one between $a,b,c$ equal to zero) I can't help myself to find some result... Further, I found some restrictions for $k,m,n$: from the expressions obtained in the system it results that, since $a,b,c\in\mathbb N$, either exactly two or none between $k,m,n$ is odd. At the moment, I can't even find an alternative approach to this... Ideas?