Fermat Last Theorem

• Feb 12th 2013, 03:58 PM
Gibo
Fermat Last Theorem
Hi,

I am not sure if I am approaching the problem below in the right way. Is there any chance that someone can check it for me,please?

Show that the equation

$\displaystyle x^2=3(y^2 + z^2) has \ no \ zero \ integer \ solution.$

I assumed by contradiction that $\displaystyle (x,y,z)$ is an integer solution to our problem and it has the smallest value of $\displaystyle z$ of any such solution.
We can also note that $\displaystyle (x, \sqrt{3}y, \sqrt{3}z)$ is in fact a Pythagoran triple as $\displaystyle 3y^2+3z^2=x^2$.
Suppose that
$\displaystyle \sqrt{3}y=2st\\ \sqrt{3}z=s^2+t^2\\x=s^2+t^2$

for some co-prime integers $\displaystyle s \ and \ t$. Let $\displaystyle s$ be even and $\displaystyle t$ be odd, but that also imlpies that either $\displaystyle s$ or $\displaystyle t$ is divisible by$\displaystyle \sqrt{3}$. Hence contradict the fact that they are integer.

Thank you
• Feb 12th 2013, 08:27 PM
johng
Re: Fermat Last Theorem
If I were grading your answer, I wouldn't accept it. What do you mean by $\displaystyle \sqrt3$ divides an integer?

Suppose x, y and z satisfy the equation. As you point out, we can assume gcd(x,y,z)=1. Clearly 3 divides x, so write $\displaystyle x=3x_0$, and the equation becomes $\displaystyle 3x_0x_0=y^2+z^2$. So 3 divides $\displaystyle y^2+z^2$. Since gcd(x,y,z)=1, 3 does not divide both $\displaystyle y^2$ and $\displaystyle z^2$. So 3 can divide neither. But then both $\displaystyle y^2$ and $\displaystyle z^2$ must be 1 mod 3, and so $\displaystyle y^2+z^2$ is 2 mod 3, and not 0. Contradiction.