(a) is simple bashing, so I will do (b). I claim that $\boxed{k=4}$. The proof is composed of two parts - showing $k \le 3$ fails and $k=4$ works. First, I show that $k\le 3$ fails. Key Lemma: $1^k+2^k+ \cdots + (p-1)^k \equiv 0 \pmod{p}$ if $k \not\equiv 0 \pmod{p-1}$. Now this gives $$1^k+2^k+3^k+4^k \equiv 1^k+2^k+3^k+4^k+5^k \pmod{5}$$so $\sum_{i=1}^n i^k$ does not form a complete residue set in $\pmod{5}$ which gives our conclusion. Done. Now, we show $k=4$ works. We will show more - $\sum_{i=1}^n i^4$ form a complete residue set in $\pmod{20^l}$ for all $l$. Check that $\sum_{i=1}^n i^5 = \frac{6n^5+15n^4+10n^2-n}{30}$, and that this forms a complete residue set in $\pmod{20}$. We go by induction. The base case $l=1$ is done by brute-force. Now we go from $l$ to $l+1$. Set $X \equiv \frac{6n^5+15n^4+10n^2-n}{30} \pmod{20^l}$, with $0 \le X < 20^l$. We will now find a suitable $m_i$ for $0 \le i \le 19$ such that $$X+ i \cdot 20^l \equiv \frac{6m^5_i + 15m^4_i + 10m^2_i-m_i}{30} \pmod{20^{l+1}}$$For simplicity, denote $f(n)=6n^5+15n^4+10n^2-n$. One can find, via brute-force expansion, $$f(n+60^l \cdot t) \equiv f(n) \pmod{3}$$$$f(n+60^l \cdot t) \equiv f(n)+(30n^4+60n^3+20n-1) \cdot 60^l \cdot t \pmod{2^{2l+3}}$$$$f(n+60^l \cdot t) \equiv f(n)+(30n^4+60n^3+20n-1) \cdot 60^l \cdot t \pmod{5^{l+2}}$$so in conclusion, we have $$f(n+60^l \cdot t) \equiv f(n)+(30n^4+60n^3+20n-1) \cdot 60^l \cdot t \pmod{30 \cdot 20^{l+1}}$$we are searching for $t$ such that $$f(n+60^l \cdot t) \equiv f(n) + i \cdot 30 \cdot 20^l \pmod{30 \cdot 20^{l+1}}$$This reduces to finding a $t$ that $$(30n^4+60n^3+20n-1) \cdot 60^l \cdot t \equiv i \cdot 30 \cdot 20^l \pmod{30 \cdot 20^{l+1}}$$or $$(30n^4+60n^3+20n-1) \cdot 3^{l-1} \cdot t \equiv 10i \pmod{200}$$Take $t=10t'$, then we have $$(30n^4+60n^3+20n-1) \cdot 3^{l-1} \cdot t' \equiv i \pmod{20}$$and as $(30n^4+60n^3+20n-1,20)=1$ and $(3^{l-1},20)=1$, for each $i$, we can find a suitable $t'$. Therefore, our induction is complete and $k=4$ works. Done. $\blacksquare$

phuong wrote: A positive interger number $k$ is called “$t-m$”-property if forall positive interger number $a$, there exists a positive integer number $n$ such that ${{1}^{k}}+{{2}^{k}}+{{3}^{k}}+...+{{n}^{k}} \equiv a (\bmod m).$ a) Find all positive integer numbers $k$ which has $t-20$-property. b) Find smallest positive integer number $k$ which has $t-{{20}^{15}}$-property. any solution part a) ?

Jjesus wrote: phuong wrote: A positive interger number $k$ is called “$t-m$”-property if forall positive interger number $a$, there exists a positive integer number $n$ such that ${{1}^{k}}+{{2}^{k}}+{{3}^{k}}+...+{{n}^{k}} \equiv a (\bmod m).$ a) Find all positive integer numbers $k$ which has $t-20$-property. b) Find smallest positive integer number $k$ which has $t-{{20}^{15}}$-property. any solution part a) ? Part a is very simple, first by CRT it is sufficient to check for $t-4$ and $t-5$ property. For $k = 1$, you obtain a quadratic function so there will be quadratic non-residue. Same for $k = 3$ since the resulting sum is a square. For $k = 2$, it is suffice to note that $1^2 + 2^2 \equiv 0 (mod 5)$ and $1^2 + 2^2 + 3^2 + 4^2 \equiv 0 (mod 5)$ so it doesn't work either. $k = 4$ is simply applying Fermat's little theorem. Part b is motivated by Hensel's lemma, where you note that the sum is a polynomial degree $k+1$.