Solve in integers the equation $$x^2+y^2+z^2+xy+yz+zx = 2w^2.$$
A trivial solution to the equation is $x = y = z = w = 0$. We can rewrite the given equation as $$(x+y)^2+(x+z)^2+(y+z)^2 = 4w^2.$$ I then thought about doing a substitution, but didn't see how to do it without becoming computational. How can we continue?
the primitive solutions to $$ a^2 + b^2 + c^2 = d^2, $$ that is $$ \gcd(a,b,c,d) = 1, $$ are given by the quaternion norm; We must have $d$ odd and one of the others, say $a.$ as well. Then $$ a = p^2 + q^2 - r^2 - s^2, $$ $$ b = 2(-ps +qr), $$ $$ c = 2(pr+qs), $$ $$ d = p^2 + q^2 + r^2 + s^2. $$ This is with $p+q+r+s$ odd, along with $\gcd(p,q,r,s)=1.$ This formula was surely known to Euler. However, the first acceptable proof that all primitive solutions occur this way was by L. E. Dickson, about 1920.
There is a way to get the formulas above. I have not used letter $t$ yet, take $$ t = p + qi + rj + sk, $$ then $$ \bar{t}i t = ai+bj+ck. $$
Note that, for your $e^2 + f^2 + g^2 = 4 w^2,$ we must have $e,f,g$ all even. So, take $$ x+y = 2a, y+z = 2b, z+x = 2c, w = d. $$ Apparently $$ x = a-b+c, \; \; y = a+b - c, \; \; z = -a +b+c. $$ Very Heronian, and all odd.
For the curious, the integer values of $$ x^2 + y^2 + z^2 + yz + zx + xy $$ are exactly the same as the integer values of $$ u^2 + v^2 + 2 w^2, $$ that is, all positive integers except $$ 4^k (16 n + 14). $$ See ME