a^2 + b^2 = 3c^2 has no integral solutions can be proved with the algebra of odd and even:
o+o=e, e+e=e, e+o=o
oxo=o, exe=e, exo=e
Substituting (a,b,c) odd or even into original equation gives:
1) (o,o,o) → e=o, (o^2 +o^2 = 3o^2)
2) (o,e,e) → o=e
3) (e,e,0) → e=o
4) (e,e,e) → e=e
5) (o,o,e) → e=e
6) (o,e,o) → o=o
1), 2) and 3) ruled out because e=o.
4) ruled out because dividing by 2^2 (again if nec) gives one of the other cases.
5) let a=2l+1, b=2m+1, c=2n. Then,
2l^2 + 2l + 2m^2 + 2m = 6n^2 – 1 → e=o
6) let a=2l+1, b=2m+1, c=2n +1. Then,
4l^2 + 4l + 4m^2+4m = 12n^2 +12n +1 → e=o
Since that exhausts the possibilities for an integral solution, there isn’t any.
I’m going to stretch my welcome by adding these 2 short videos which give, in a simple manner, the modular arithmetic approach to solving this type of problem.
I only meant to show the links. I don't know why big pictures of the site showed up. Don't mean to hog that much space. Try again:
There, that's better. Wonder why the site showed up. Original link pasted underlined for some reason. Interesting.
Damn, did it again. When I previewed post the links came back underlined and the sites showed up. Don't understand. Oh well.
jojo7777777, you asked about Z (congruence arithmetic).
A reasonable explanation is the first few pages of:
www.math.ou.edu/~kmartin/nti/chap3.pdf
(Post 16 is still a valid elementary algebra sol to OP if you don't have to do it with congruences.)