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

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- February 18th 2013, 11:50 PMjojo7777777show 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 - February 19th 2013, 03:24 AMPazeRe: show that a^2+b^2=3c^2 has no solutions in the integers
For a second there I thought you were trying to prove Fermat's Last Theorem haha.

However your equation does have a solution, namely - February 19th 2013, 05:58 AMMathJackRe: show that a^2+b^2=3c^2 has no solutions in the integers
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

- February 19th 2013, 06:24 AMKmathRe: show that a^2+b^2=3c^2 has no solutions in the integers
- February 19th 2013, 06:35 AMMathJackRe: show that a^2+b^2=3c^2 has no solutions in the integers
I retract that statement..apologies

- February 19th 2013, 07:06 AMKmathRe: show that a^2+b^2=3c^2 has no solutions in the integers
- February 19th 2013, 11:54 AMHartlwRe: show that a^2+b^2=3c^2 has no solutions in the integers
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. - February 20th 2013, 09:54 AMjojo7777777Re: show that a^2+b^2=3c^2 has no solutions in the integers
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? - February 20th 2013, 10:25 AMHartlwRe: show that a^2+b^2=3c^2 has no solutions in the integers
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) - February 20th 2013, 03:21 PMHartlwRe: show that a^2+b^2=3c^2 has no solutions in the integers
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. - February 21st 2013, 05:05 AMHartlwRe: show that a^2+b^2=3c^2 has no solutions in the integers
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. - February 21st 2013, 06:24 AMHartlwRe: show that a^2+b^2=3c^2 has no solutions in the integers
- February 22nd 2013, 08:10 AMHartlwRe: show that a^2+b^2=3c^2 has no solutions in the integers
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 - February 22nd 2013, 10:19 AMHartlwRe: show that a^2+b^2=3c^2 has no solutions in the integers
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 - February 22nd 2013, 12:53 PMHartlwRe: show that a^2+b^2=3c^2 has no solutions in the integers
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)