Prove that there are no such positive integers $a,b,c,d$ such that $a^2 + b^2 = 3(c^2 + d^2)$ Prove that there are no positive integers a, b, c, d such that $a^2 + b^2 = 3(c^2 + d^2)$. Hint: What can you say about divisibility of a and b by 3? Look at solution with smallest possible a. 

By way of contradiction, suppose there exist a,b,c,d $\in \mathbb N$ such that $a^2 + b^2 = 3(c^2 + d^2)$.  
(Note: see that the square of a natural number is also natural and the sum of natural numbers is also natural.)
Now let $a_o , b_o , c_o , d_o $ be the smallest solution set to the equation.
Then 3|$a_o$$^2 + b_o$$^2 \Rightarrow 3 | a_o$$^2$ and $ 3|b_o$$^2$ $\Rightarrow 3|a_o$ and $3|b_o$.
$\therefore$ $a_o = 3n$ and $b_o = 3m$, where m,n $\in \mathbb N$ such that n < $a_o$ and m < $b_o$.  
Then $9(n^2 + m^2) = 3(c_o$$^2$ + $d_o$$^2)\Rightarrow 3(n^2 + m^2) = c_o$$^2 + d_o$$^2$,
Then by the similar argument above $3 | c_o$ and $ 3|d_o$.
$\therefore$ $c_o = 3r$ and $d_o = 3p$, where r,p $\in \mathbb N$ such that r < $c_o$ and p < $d_o$.  
Now see that $9(n^2 + m^2) = 27(r^2 + p^2) \Rightarrow n^2 + m^2 = 3(r^2 + p^2)$, but that means n, m, r, p are a solution set to the equation $a^2 + b^2 = 3(c^2 + d^2)$ smaller than $a_o , b_o , c_o , d_o $, but this is a contradiction since $a_o , b_o , c_o , d_o $ is the smallest solution set, hence our assumption that there exist a,b,c,d $\in \mathbb N$ such that $a^2 + b^2 = 3(c^2 + d^2)$ is false.  
