HAVE YOU EVER TRIED: 2, 3, 5, 6?

We have : \(2018.4=\left(1^2+1^2+1^2+1^2\right)\left(a^4+b^4+c^4+d^4\right)\ge\left(a^2+b^2+c^2+d^2\right)^2\) 

(Inquality Bunyakovsky)

\(\Leftrightarrow-\sqrt{2018.4}\le a^2+b^2+c^2+d^2\le\sqrt{2018.4}\)

\(\Leftrightarrow0\le a^2+b^2+c^2+d^2< 90\)

Cause \(a^4,b^4,c^4,d^4\) are square numbers 

So when they divided for 3 , Their remainder must be 1 or 0

<=> \(\left(a^4+b^4+c^4+d^4\right)\equiv1\left(mod3\right)\)

or \(\left(a^4+b^4+c^4+d^4\right)\equiv0\left(mod3\right)\)

But \(2018\equiv2\left(mod3\right)\)

So there isn't any root of (a;b;c;d) that satisfy the equation

It also means that we can't calculate the value of \(a^2+b^2+c^2+d^2\)

But there is only one value for a² + b² + c² + d², can you find that?