For a second there I thought you were trying to prove Fermat's Last Theorem haha.
However your equation does have a solution, namely
I looked at the link above but couldn't follow it. Somewhere I have a short paperback on number theory. If I can find it, I may take another crack at it.
The furthest I could get was that a^2 + b^2 couldn't equal I^2 because that would make sqrt(3) rational. But couldn't get any further. (but even so you have to prove sqrt(3) is irrational)
Assume a^2 + b^2 = 3c^2 has integral solutions.
divide by 3c^2 to get
a’^2 + b’^2 = 1, which has rational sols r, so that
a’ = a/sqrt3c = r →
sqrt3 = a/rc impossible
therefore, integral sols to a^2 + b^2 = 3c^2 do not exist.
EDIT: previous post is OK. The fact that a'^2 + b'^2 = 1 HAS to have a rational solution does lead to a contradiction.
To appreciate this solution, check out the link in post #6.
jojo777777: Thanks for the thanks.
The good news: My solution was short, simple, and transparent.
The bad news: It was wrong. I pissed in your ear and told you it was raining- my apologies.
Integral sols of x^2 + y^2 = 3z^2 are rational sols of x^2 + y^2 = 3.
x^2 + y^2 = 1 has rational points (3/5,4/5), but they don’t work for r =sqrt3. Tried looking at trig of circle, no use.
Looks like one is stuck with learning modular arithmetic to understand previous link.
You might find it comforting to look at someone else’s struggle with this problem (the proof is wrong):
[number theory] prove that x^2+y^2=3 has no rational points
definition: a congruent b mod m iff m|(b-a) or a-b=mc for some c.
If a,b, c are integers, the first step of link in post 6 says:
a^2 + b^2 congruent 3c^2 mod 4, or a^2 + b^2 –3c^2 = 4d for some d.
Where does that come from, ie, how do you know a^2 + b^2 –3c^2 is divisible by 4?
EDIT: If a is even, a=2k and a^2 = 4k^2. But if a =2k+1, i guess if you look at the different possibilities for odd and even, it works out. Will ck.
A very simple solution is given in:
The first step of link in post 6 says, in effect, that a cong a mod 4, which is correct but trivial and misleading, since a cong a for any integer. I’ll have to educate myself on modular arithmetic to judge whether it subsequently makes sense.
References for modular arithmetic:
google: modular arithmetic (wiki), and Congruence mod n (Harvard video)
Among other things, they will explain Z (congruence classes)