Is 2018 special because of these properties? I discovered that: $$2018=(6^2)^2+(5^2)^2+(3^2)^2+(2^2)^2$$
We also have:  $$13^2+43^2=2018$$
And we have:
$$2018=44^2+9^2+1^2$$
I somehow tend to believe that there could be a finite number of these numbers that are sum of two squares, three squares and four fourth powers.
So we have a system of three Diophantine equations:
$$n=a^2+b^2$$
and $$n=c^2+d^2+e^2$$
and $$n=f^4+g^4+h^4+i^4$$
where, $n,a,b,c,d,e,f,g,h,i \in \mathbb N$.

Is there a finite number of these numbers?

Edit : Also, it is $$2018=35^2+26^2+8^2+7^2+2^2$$ a sum of five squares.
And of $$2018=11^3+7^3+7^3+1^3$$ four cubes.

One infinite set is $2018k^4$ for any natural $k$.  I strongly suspect that there are plenty more.  Numbers that are a sum of three squares are very common.  Numbers that are a sum of two squares are not so rare, so I would just start picking sums of four cubes and try to satisfy the other two.  Another example is $$1^4+2^4+3^4+6^4=1394=2\cdot 17 \cdot 41 \equiv 4\pmod 8$$
so is a sum of two and three squares.
