<s>[INST] 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.
 [/INST] There are an infinite number of such integers.
One way to show this is to start from the fact that there are an infinite number of Pythagorean triples $a^2 + b^2 = c^2$.  Given any such triple, let $N$ be given by:
$$N = (ab)^4 + (bc)^4 + (ac)^4 + c^4$$
By an identity of Fauquembergue (1) we have (given $a^2 + b^2 = c^2$):
$$(ab)^4 + (bc)^4 + (ac)^4 = (a^4 + a^2b^2 + b^4)^2$$
Hence:
$$N = (a^4 + a^2b^2 + b^4)^2 + (c^2)^2$$
Furthermore:
$$(c^2)^2 = (a^2 + b^2)^2 = a^4 + 2a^2b^2 + b^4 = (a^2 - b^2)^2 + (2ab)^2$$
Hence:
$$N = (a^4 + a^2b^2 + b^4)^2 + (a^2 - b^2)^2 + (2ab)^2$$
Reference:
1) Dickson L E History of the Theory of Numbers Vol 2 Ch XXII p 658
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