Enjoying 2018 while it lasts

Since 1009 is prime, Fermat tells us that $k^{1009}\equiv k \pmod{1009}$ for all $k$ . Squaring gives $k^{2018}\equiv k^2 \pmod{1009}$ . This congruence holds $\bmod{2018}$ as well since $k^{2018}$ and $k^2$ have the same parity. We are asked to find the smallest positive integer $N$ such that $\sum_{k=1}^{N}k^2$ is divisible by 2018. Now $\sum_{k=1}^{N}k^2=\frac{N(N+1)(2N+1)}{6}$ . The smallest $N$ such that this sum is divisible by 1009 is $N=504$ , when $2N+1=1009$ . We check that $\sum_{k=1}^{504}k^2=\frac{504\times 505 \times1009}{6}$ is even, so that the answer is $N=\boxed{504}$ .

Since the prime factorization of 2018 is 2×1009, we can see that $x^{2018} = (x^{1009})^2 ≡ x^2$ (mod 2018), so N is also the smallest positive integer N such that $\sum_{k=1}^N{k^2}$ is divisible by 2018. This sum is equal to $\frac{N(N+1)(2N+1)}{6}$ . In order for that to be divisible by 2018 it needs to contain a factor 1009, which can either be a factor of N, of N+1 or of 2N+1. The smallest possibilty is 2N+1=1009 => N= 504. We try it out: $\frac{2×504^3+ 3×504^2+ 504}{6}=21210×2018$ , showing that this is a successful try, $\fbox{N=504}$ .

By using python we obtain as output $\boxed{504}$ .

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a,s = 1,1
while s%2018!=0:
    a+=1
    s+=a**2018
print a

Using $Mathematica$

n = 1; While[Mod[Sum[k^2018, {k, n++}], 2018] != 0]; n - 1

returns 504

Use the eulers theorem

A^(phi{n})=1 mod n ... So Phi (2018)=1008... So A^1008=1mod 2018... Squaring the congruence... A^2016=1 mod 2018... Or A^2018=A^2 mod 2018... So the sequence is divisible by 2018 iff ... 1^2 +2^2...........N^2= 0 mod 2018..

So [N][N+1][2N+1]/6 is=0 mod 2018...

As largest prime factor of 2018 is 1009.. Let 2N+1=1009..

Or N=504... Note::phi stands for the eulers totient function

Tried WolframAlpha script:

"Sum (x^2018 from x=0 to N) mod 2018"

for

N= 1009 (Answer=1009),

N=1008 (Answer=0),

and

N= 1008/2=504 (Answer=0),

Therefore,

Answer=504