Exercise 0.3.8 (Equation $a^2 + b^2 = 3c^2$)

Question

Prove that the equation $a^2 + b^2 = 3c^2$ has no solutions in nonzero integers $a$, $b$ and $c$.

[Consider the equation $\mod 4$ as in the previous two exercise and show that $a$, $b$ and $c$ would all have to be divisible by $2$. Then each of $a^2$, $b^2$ and $c^2$ has a factor of $4$ and by dividing through by $4$ show that there would be a smaller set of solutions to the original equation. Iterate to reach a contradiction.]

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

This is the Fermat's method of infinite descent. I write this argument in the following manner. 
\begin{proof} 
Suppose for the sake of contradiction that 
$$a^2 + b^2 = 3 c^2,\qquad 	ext{where} \qquad (a,b,c) \in \Z^3,\ (a,b,c) 
e (0,0,0).$$
Let $d = \mathrm{g.c.d.}(a,b,c)$. Since $(a,b,c) 
e (0,0,0)$, then $d
e 0$. Put $A =a/d,\ B = b/d, C = c/d$. Then $ \mathrm{g.c.d.}(A,B,C) = 1$, and
$$A^2 + B^2 = 3 C^2, \qquad 	ext{where} \qquad (A,B,C) \in \Z^3,\qquad \mathrm{g.c.d.}(A,B,C) = 1.$$
Since $C^2  \equiv 0 	ext{ or } 1 \pmod 4$, then $3C^2  \equiv 0 	ext{ or } 3 \pmod 4$, therefore
$$A^2 + B^2 \equiv 0 	ext{ or } 3 \pmod 4.$$
If $A$ and $B$ have distinct parities, then $A^2 + B^2 \equiv 1 \pmod 4$, which is false, so $A$ and $B$ have same parity. If $A$ and $B$ are both odd, then $A^2 + B^2 \equiv 2 \pmod 4$, which is also false, so $A$ and $B$ are both even. Then $3C^2$ is even, so $C$ is even. This shows that $2$ divides $1 =\mathrm{g.c.d.}(A,B,C)$. This is a contradiction, because $1 
mid 2$.

In conclusion, for all $(a,b,c) \in \Z^3$,
$$a^2 + b^2 = 3c^2 \Rightarrow a= b =c = 0.$$
 The equation $a^2 + b^2 = 3c^2$ has no integer solution, except $(0,0,0)$.
\end{proof}

Exercise 0.3.8 (Equation $a^2 + b^2 = 3c^2$)

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Exercise 0.3.8 (Equation $a^2 + b^2 = 3c^2$)

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