<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] Sum of 4 distinct fourth powers: 
$$2018 =2^4 + 3^4 + 5^4 + 6^4$$
as well as sum of 12 consecutive squares: 
$$2018 = 7^2 + 8^2 + 9^2 + 10^2 + 11^2 + 12^2 + 13^2 + 14^2 + 15^2 + 16^2 + 17^2 + 18^2$$
and "smallest number equal to the product of two primes which is also equal to the sum of 33 distinct primes": http://oeis.org/A102238
and the third least k > 0 such that the nextprime(k$\times$primorial(n)) - k$\times$primorial(n) is composite: http://oeis.org/A071771 
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