Thread: show that a^2+b^2=3c^2 has no solutions in the integers

How I'm showing that the equation a^2+b^2=3c^2 has no solutions (other than the trivial solution (a,b,c)=(0,0,0)) in the integers?
Thanking you in anticipation

For a second there I thought you were trying to prove Fermat's Last Theorem haha.

However your equation does have a solution, namely

Root 3 is irrational so any number divided by this is irrational unless divided into itself. There is no sum of two squares which = 3, therefore there is no natural answers

Originally Posted by MathJack

Root 3 is irrational so any number divided by this is irrational unless divided into itself.

far away from the original question, why this is true?

I retract that statement..apologies

take a look

http://hans.math.upenn.edu/~shonkwil.../502/502_1.pdf

a^2 + b^2 = 2c^2 + c^2
c^2 = (a-b)^2
c = a-b

EDIT: Sorry about that. I made a stupid mistake.

Thank you very much!!!!
Sorry for asking what does Z/4Z exactly represent, in the attached file?
Hartlw, does your answer suppose to show a contradiction?

Originally Posted by jojo7777777

Thank you very much!!!!
Sorry for asking what does Z/4Z exactly represent, in the attached file?
Hartlw, does your answer suppose to show a contradiction?

My answer was a mistake- my apologies.

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.

Originally Posted by Hartlw

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.

I add this post for the benefit of those who may have missed the important late edit to my previous post, which confirms the above sol is OK.

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:

www.qbyte.org/puzzles/p103ss.html

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)