Skip to main content
MathOverflow

Return to Answer

added 45 characters in body
Source Link
M.S
M.S
  • 29
  • 4

we can write: $\displaystyle\int_{0}^{m}{x^m}dx\le\sum_{n=1}^{m}{n^m} \le\int_{0}^{m+1}{x^m}dx$

so $\displaystyle\frac{m^{m+1}-(m+1)-m(m+1)k}{m+1}\le\sum_{n=1}^{m}{n^m}-1-mk-\le\frac{(m+1)^{m+1}-(m+1)-m(m+1)k}{m+1}$ and then $\displaystyle\frac{m^{m+1}-(m+1)-m(m+1)k_1}{m+1}\le\sum_{n=1}^{m}{n^m}-1-mk-\le\frac{(m+1)^{m+1}-(m+1)-m(m+1)k_2}{m+1}$ in which assume that $k_2\le k$ and$k\le k_1$,

but since $(m+1)^{m+1}\equiv{m+1}\pmod{m}$ and also $m^{m+1}\equiv m\pmod{m}$ so the

central statement should be zero, or $\displaystyle\sum_{n=1}^{m}{n^m}\equiv 1\bmod m$, when the first

congruence holds for the numbers 1,2,6,42,1806, second one holds for these numbers.

note: if $m$ be an even number so $m^{m}-1$ is divided to $m+1$ then $(m+1)^{m+1}\equiv{m+1}\pmod{m(m+1)}$ and $m^{m+1}\equiv m\pmod{m(m+1)}$ the above inequalities is hold for some $m$

we can write: $\displaystyle\int_{0}^{m}{x^m}dx\le\sum_{n=1}^{m}{n^m} \le\int_{0}^{m+1}{x^m}dx$

so $\displaystyle\frac{m^{m+1}-(m+1)-m(m+1)k}{m+1}\le\sum_{n=1}^{m}{n^m}-1-mk-\le\frac{(m+1)^{m+1}-(m+1)-m(m+1)k}{m+1}$ and then $\displaystyle\frac{m^{m+1}-(m+1)-m(m+1)k_1}{m+1}\le\sum_{n=1}^{m}{n^m}-1-mk-\le\frac{(m+1)^{m+1}-(m+1)-m(m+1)k_2}{m+1}$ in which assume that $k_2\le k$ and$k\le k_1$,

but since $(m+1)^{m+1}\equiv{m+1}\pmod{m}$ and also $m^{m+1}\equiv m\pmod{m}$ so the

central statement should be zero, or $\displaystyle\sum_{n=1}^{m}{n^m}\equiv 1\bmod m$, when the first

congruence holds for the numbers 1,2,6,42,1806, second one holds for these numbers.

note: if $m$ be an even number so $m^{m}-1$ is divided to $m+1$ then $(m+1)^{m+1}\equiv{m+1}\pmod{m(m+1)}$ and $m^{m+1}\equiv m\pmod{m(m+1)}$

we can write: $\displaystyle\int_{0}^{m}{x^m}dx\le\sum_{n=1}^{m}{n^m} \le\int_{0}^{m+1}{x^m}dx$

so $\displaystyle\frac{m^{m+1}-(m+1)-m(m+1)k}{m+1}\le\sum_{n=1}^{m}{n^m}-1-mk-\le\frac{(m+1)^{m+1}-(m+1)-m(m+1)k}{m+1}$ and then $\displaystyle\frac{m^{m+1}-(m+1)-m(m+1)k_1}{m+1}\le\sum_{n=1}^{m}{n^m}-1-mk-\le\frac{(m+1)^{m+1}-(m+1)-m(m+1)k_2}{m+1}$ in which assume that $k_2\le k$ and$k\le k_1$,

but since $(m+1)^{m+1}\equiv{m+1}\pmod{m}$ and also $m^{m+1}\equiv m\pmod{m}$ so the

central statement should be zero, or $\displaystyle\sum_{n=1}^{m}{n^m}\equiv 1\bmod m$, when the first

congruence holds for the numbers 1,2,6,42,1806, second one holds for these numbers.

note: if $m$ be an even number so $m^{m}-1$ is divided to $m+1$ then $(m+1)^{m+1}\equiv{m+1}\pmod{m(m+1)}$ and $m^{m+1}\equiv m\pmod{m(m+1)}$ the above inequalities is hold for some $m$

added 63 characters in body; added 81 characters in body
Source Link
M.S
M.S
  • 29
  • 4

we can write: $\displaystyle\int_{0}^{m}{x^m}dx\le\sum_{n=1}^{m}{n^m} \le\int_{0}^{m+1}{x^m}dx$

so $\displaystyle\frac{m^{m+1}-(m+1)-m(m+1)k}{m+1}\le\sum_{n=1}^{m}{n^m}-1-mk-\le\frac{(m+1)^{m+1}-(m+1)-m(m+1)k}{m+1}$ and then $\displaystyle\frac{m^{m+1}-(m+1)-m(m+1)k_1}{m+1}\le\sum_{n=1}^{m}{n^m}-1-mk-\le\frac{(m+1)^{m+1}-(m+1)-m(m+1)k_2}{m+1}$ in which assume that $k_2\le k$ and$k\le k_1$,

but since $(m+1)^{m+1}\equiv{m+1}\pmod{m}$ and also $m^{m+1}\equiv m\pmod{m}$ so the

central statement should be zero, or $\displaystyle\sum_{n=1}^{m}{n^m}\equiv 1\bmod m$, when the first

congruence holds for the numbers 1,2,6,42,1806, second one holds for these numbers.

note: if $m$ be an even number so $m^{m}-1$ is divided to $m+1$ then $(m+1)^{m+1}\equiv{m+1}\pmod{m(m+1)}$ and $m^{m+1}\equiv m\pmod{m(m+1)}$

we can write: $\displaystyle\int_{0}^{m}{x^m}dx\le\sum_{n=1}^{m}{n^m} \le\int_{0}^{m+1}{x^m}dx$

so $\displaystyle\frac{m^{m+1}-(m+1)-m(m+1)k}{m+1}\le\sum_{n=1}^{m}{n^m}-1-mk-\le\frac{(m+1)^{m+1}-(m+1)-m(m+1)k}{m+1}$ and then $\displaystyle\frac{m^{m+1}-(m+1)-m(m+1)k_1}{m+1}\le\sum_{n=1}^{m}{n^m}-1-mk-\le\frac{(m+1)^{m+1}-(m+1)-m(m+1)k_2}{m+1}$ in which assume that $k_2\le k$ and$k\le k_1$,

but since $(m+1)^{m+1}\equiv{m+1}\pmod{m}$ and also $m^{m+1}\equiv m\pmod{m}$ so the

central statement should be zero, or $\displaystyle\sum_{n=1}^{m}{n^m}\equiv 1\bmod m$, when the first

congruence holds for the numbers 1,2,6,42,1806, second one holds for these numbers

we can write: $\displaystyle\int_{0}^{m}{x^m}dx\le\sum_{n=1}^{m}{n^m} \le\int_{0}^{m+1}{x^m}dx$

so $\displaystyle\frac{m^{m+1}-(m+1)-m(m+1)k}{m+1}\le\sum_{n=1}^{m}{n^m}-1-mk-\le\frac{(m+1)^{m+1}-(m+1)-m(m+1)k}{m+1}$ and then $\displaystyle\frac{m^{m+1}-(m+1)-m(m+1)k_1}{m+1}\le\sum_{n=1}^{m}{n^m}-1-mk-\le\frac{(m+1)^{m+1}-(m+1)-m(m+1)k_2}{m+1}$ in which assume that $k_2\le k$ and$k\le k_1$,

but since $(m+1)^{m+1}\equiv{m+1}\pmod{m}$ and also $m^{m+1}\equiv m\pmod{m}$ so the

central statement should be zero, or $\displaystyle\sum_{n=1}^{m}{n^m}\equiv 1\bmod m$, when the first

congruence holds for the numbers 1,2,6,42,1806, second one holds for these numbers.

note: if $m$ be an even number so $m^{m}-1$ is divided to $m+1$ then $(m+1)^{m+1}\equiv{m+1}\pmod{m(m+1)}$ and $m^{m+1}\equiv m\pmod{m(m+1)}$

added 180 characters in body; edited body; edited body
Source Link
M.S
M.S
  • 29
  • 4

we can write: $\displaystyle\int_{0}^{m}{x^m}dx\le\sum_{n=1}^{m}{n^m} \le\int_{0}^{m+1}{x^m}dx$

so $\displaystyle\frac{m^{m+1}-(m+1)-m(m+1)k}{m+1}\le\sum_{n=1}^{m}{n^m}-1-mk\le\frac{(m+1)^{m+1}-(m+1)-m(m+1)k}{m+1}$$\displaystyle\frac{m^{m+1}-(m+1)-m(m+1)k}{m+1}\le\sum_{n=1}^{m}{n^m}-1-mk-\le\frac{(m+1)^{m+1}-(m+1)-m(m+1)k}{m+1}$ and then $\displaystyle\frac{m^{m+1}-(m+1)-m(m+1)k_1}{m+1}\le\sum_{n=1}^{m}{n^m}-1-mk-\le\frac{(m+1)^{m+1}-(m+1)-m(m+1)k_2}{m+1}$ in which assume that $k_2\le k$ and$k\le k_1$,

but since $(m+1)^{m+1}\equiv{m+1}\pmod{m}$ and also $m^{m+1}\equiv m\pmod{m}$ so the

central statement should be zero, or $\displaystyle\sum_{n=1}^{m}{n^m}\equiv 1\bmod m$, when the first

congruence holds for the numbers 1,2,6,42,1806, second one holds for these numbers

we can write: $\displaystyle\int_{0}^{m}{x^m}dx\le\sum_{n=1}^{m}{n^m} \le\int_{0}^{m+1}{x^m}dx$

so $\displaystyle\frac{m^{m+1}-(m+1)-m(m+1)k}{m+1}\le\sum_{n=1}^{m}{n^m}-1-mk\le\frac{(m+1)^{m+1}-(m+1)-m(m+1)k}{m+1}$

but since $(m+1)^{m+1}\equiv{m+1}\pmod{m}$ and also $m^{m+1}\equiv m\pmod{m}$ so the

central statement should be zero, or $\displaystyle\sum_{n=1}^{m}{n^m}\equiv 1\bmod m$, when the first

congruence holds for the numbers 1,2,6,42,1806, second one holds for these numbers

we can write: $\displaystyle\int_{0}^{m}{x^m}dx\le\sum_{n=1}^{m}{n^m} \le\int_{0}^{m+1}{x^m}dx$

so $\displaystyle\frac{m^{m+1}-(m+1)-m(m+1)k}{m+1}\le\sum_{n=1}^{m}{n^m}-1-mk-\le\frac{(m+1)^{m+1}-(m+1)-m(m+1)k}{m+1}$ and then $\displaystyle\frac{m^{m+1}-(m+1)-m(m+1)k_1}{m+1}\le\sum_{n=1}^{m}{n^m}-1-mk-\le\frac{(m+1)^{m+1}-(m+1)-m(m+1)k_2}{m+1}$ in which assume that $k_2\le k$ and$k\le k_1$,

but since $(m+1)^{m+1}\equiv{m+1}\pmod{m}$ and also $m^{m+1}\equiv m\pmod{m}$ so the

central statement should be zero, or $\displaystyle\sum_{n=1}^{m}{n^m}\equiv 1\bmod m$, when the first

congruence holds for the numbers 1,2,6,42,1806, second one holds for these numbers

added 15 characters in body
Source Link
M.S
M.S
  • 29
  • 4
added 19 characters in body
Source Link
M.S
M.S
  • 29
  • 4
improved formatting, corrected spelling, fixed grammar
Source Link
Gerry Myerson
Gerry Myerson
  • 39.9k
  • 10
  • 186
  • 247
deleted 29 characters in body
Source Link
M.S
M.S
  • 29
  • 4
Source Link
M.S
M.S
  • 29
  • 4