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. 
 A: Here are some hints for you regarding this question:
Use the fact that $2$ isn't quadratic residue modulo $3$. The RHS is divisible by ... $a,b$ are of the form ...
Then assume that $(a_0,b_0,c_0,d_0)$ is the smallest solution, using the things we already stated prove there's another smaller solutions $(a_1,b_1,c_1,d_1)$, which means that ... 
This method is called infinite descent, you can learn something more if you just put some little bit more effort.
The places with three dots are left for you to fill them.
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.  
A: Well, $A^2+B^2=1,0$ ( mod 4) so it couldn't be $A^2+B^2=4k+3$. But (4k+1)$(4k+3)=16k^2+16k+3=4(4k^2+4k)+3=4r+3$, while (4k+3)$(4k+3)=16k^2+24k+8+1=4(4k^2+6k+2)+1=4s+1$. So if a prime $p|A^2+B^2$ and p=4k+3, $p^2|A^2+B^2!$ But the exponent of number 3 is odd while it should be even. Hence there no positive integers A,B,C,D such that $A^2+B^2=3(C^2+D^2)$.
