Algebra Solutions: Prove that a given Diophantine equation has no nontrivial solutions

Prove that the equation $a^2 + b^2 = 3c^2$ has no solutions in nonzero integers $a$ , $b$ , and $c$ .

Let $S$ be the set of all nonzero solutions $(a,b,c)$ of the equation $a^2 + b^2 = 3c^2$ . Consider now the set $A = \{ a \ |\ (a,b,c) \in S \}$ ; this is a nonempty set of positive integers, and so by the Well Ordering Principle $A$ has a minimal element $a_0$ . Then let $(a_0,b_0,c_0)$ be a solution. We now reduce our arithmetic mod 4. By the previous exercises, $a_0^2 + b_0^2$ is either $0$ , $1$ , or $2$ mod 4, and $c_0^2$ is either $0$ or $1$ mod 4. But if $c_0^2$ is 1 mod 4, then we have $a_0^2 + b_0^2 = 3$ mod 4, a contradiction. So $c^2$ is 0 mod 4, and thus 2 divides $c_0$ . Moreover, we have $a_0^2 + b_0^2 = 0$ mod 4, and by a previous example, the only way this can happen is if $a^2 = b^2 = 0$ mod 4. So 2 divides $a_0$ and $b_0$ .
Now we can construct a new integer solution $(a_0/2, b_0/2, c_0/2)$ of $a^2 + b^2 = 3c^2$ . However, $a_0/2$ is strictly less than $a_0$ , violating the minimality of $a_0$ . Hence we have a contradiction, so $S$ is empty.

Algebra Solutions

Prove that a given Diophantine equation has no nontrivial solutions

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