We can use this property of squares to reach the solution
The Square of a natural number other than one is either a multiple of 3 or exceeds a multiple of 3 by 1.
In other words, a perfect square leaves remainder 0 or 1 on division by 3.
Given equation is
$x^2+y^2=3(z^2+t^2)$ This implies $x^2+y^2$ would be a multiple of $3.$
There are $4$ possibilities for $x^2$ and $y^2$
- $x^2=3a$ and $y^2=3b+1$ so $x^2+y^2= 3(a+b)+1$ won't be multiple of 3 so reject such type of x and y
- $x^2=3a+1$ and $y^2=3b+1$ so $x^2+y^2=3(a+b)+2$ won't be multiple of 3 so reject such type of x and y
- $x^2=3a+1$ and $y^2=3b$ // same as possibility 1) so reject
- $x^2=3a$ and $y^2=3b$ so $x^2+y^2=3(a+b)$ // multiple of 3 so x and y would be of such type
Thus, here we can simplify the equation to $3(a+b)=3(z^2+t^2)$
$\implies a+b=z^2+t^2$ // if we observe square of numbers then we can see that $a+b$ would also be a multiple of $3$ and more clearly it would be like $3\times p$ where $p$ is not multiple of $3.$
As the previous $4$ possibilities there are again $4$ possibilities for $z^2+t^2.$ Now as $a+b$ is a multiple of $3,$ $z^2+t^2$ must also be a multiple of $3.$ So, $z^2=3c$ and $t^2=3d.$
Let $a+b=3p$ and $3p=3(c+d)$
so eventually $p=c+d.$ According to my last claim $c+d$ must also be multiple of $3$ but $p$ won't be a multiple of three so this $p$ can never be equal to $c+d.$
Hence No Solution. Option E
