Mixing
Sums of first $n$ $k$th powers which are perfect powers (as polynomials in $n$)
$begingroup$
This is likely a very basic question, but I have not been able to find an answer (or even someone posing this question), so here goes.
Let $P_k(n) = sum_{m=1}^n m^k$. It is well-known that
$$displaystyle P_3(n) = left(frac{n(n+1)}{2}right)^2.$$
In particular, $P_3(n)$ is a perfect square as a polynomial in $n$. Is $k = 3$ the only $k$ for which this is the case? That is, does there exist $k geq 4$ such that $P_k(n)$ is a perfect $ell$-th power for some $ell geq 2$?
number-theory
asked Jan 17 at 3:01
syxiaosyxiao
808816
$endgroup$
add a comment |
$begingroup$
This is likely a very basic question, but I have not been able to find an answer (or even someone posing this question), so here goes.
Let $P_k(n) = sum_{m=1}^n m^k$. It is well-known that
$$displaystyle P_3(n) = left(frac{n(n+1)}{2}right)^2.$$
In particular, $P_3(n)$ is a perfect square as a polynomial in $n$. Is $k = 3$ the only $k$ for which this is the case? That is, does there exist $k geq 4$ such that $P_k(n)$ is a perfect $ell$-th power for some $ell geq 2$?
number-theory
asked Jan 17 at 3:01
syxiaosyxiao
808816
$endgroup$
1
$begingroup$
I do not think so but it would nice since it would be an extension of Nicomachus's theorem.
$endgroup$
– Claude Leibovici
Jan 17 at 3:20
1
$begingroup$
Bernoulli gave a way to obtain general algebraic expressions for what you call $P_k(n)$ for any $k$.
$endgroup$
– Keith Backman
Jan 17 at 3:40
add a comment |
$begingroup$
This is likely a very basic question, but I have not been able to find an answer (or even someone posing this question), so here goes.
Let $P_k(n) = sum_{m=1}^n m^k$. It is well-known that
$$displaystyle P_3(n) = left(frac{n(n+1)}{2}right)^2.$$
In particular, $P_3(n)$ is a perfect square as a polynomial in $n$. Is $k = 3$ the only $k$ for which this is the case? That is, does there exist $k geq 4$ such that $P_k(n)$ is a perfect $ell$-th power for some $ell geq 2$?
number-theory
asked Jan 17 at 3:01
syxiaosyxiao
808816
$endgroup$
This is likely a very basic question, but I have not been able to find an answer (or even someone posing this question), so here goes.
Let $P_k(n) = sum_{m=1}^n m^k$. It is well-known that
$$displaystyle P_3(n) = left(frac{n(n+1)}{2}right)^2.$$
In particular, $P_3(n)$ is a perfect square as a polynomial in $n$. Is $k = 3$ the only $k$ for which this is the case? That is, does there exist $k geq 4$ such that $P_k(n)$ is a perfect $ell$-th power for some $ell geq 2$?
number-theory
number-theory
asked Jan 17 at 3:01
syxiaosyxiao
808816
asked Jan 17 at 3:01
syxiaosyxiao
808816
asked Jan 17 at 3:01
syxiaosyxiao
808816
asked Jan 17 at 3:01
syxiaosyxiao
808816
asked Jan 17 at 3:01
syxiaosyxiao
808816
808816
1
$begingroup$
I do not think so but it would nice since it would be an extension of Nicomachus's theorem.
$endgroup$
– Claude Leibovici
Jan 17 at 3:20
1
$begingroup$
Bernoulli gave a way to obtain general algebraic expressions for what you call $P_k(n)$ for any $k$.
$endgroup$
– Keith Backman
Jan 17 at 3:40
add a comment |
1
$begingroup$
I do not think so but it would nice since it would be an extension of Nicomachus's theorem.
$endgroup$
– Claude Leibovici
Jan 17 at 3:20
1
$begingroup$
Bernoulli gave a way to obtain general algebraic expressions for what you call $P_k(n)$ for any $k$.
$endgroup$
– Keith Backman
Jan 17 at 3:40
1
1
$begingroup$
I do not think so but it would nice since it would be an extension of Nicomachus's theorem.
$endgroup$
– Claude Leibovici
Jan 17 at 3:20
$begingroup$
I do not think so but it would nice since it would be an extension of Nicomachus's theorem.
$endgroup$
– Claude Leibovici
Jan 17 at 3:20
1
1
$begingroup$
Bernoulli gave a way to obtain general algebraic expressions for what you call $P_k(n)$ for any $k$.
$endgroup$
– Keith Backman
Jan 17 at 3:40
$begingroup$
Bernoulli gave a way to obtain general algebraic expressions for what you call $P_k(n)$ for any $k$.
$endgroup$
– Keith Backman
Jan 17 at 3:40
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There is no other $k$.
Faulhaber's formula for $F_k(x)$ has rational coefficients and leading term $1/(k+1)$. If $F_k(x)$ is the square of a polynomial, we have
$$ F_k(x) = frac{g(x)^2}{k+1}$$
where $g(x)^2$ is monic with rational coefficients. Now $g(x)$, being the square root of a monic polynomial with rational coefficients, has rational coefficients (in fact the square root of a monic polynomial over the rationals can be written as a Laurent series over the rationals; it is a polynomial if the terms in negative powers of $x$ are all $0$).
In particular,
$$ 1 + 2^k = frac{F_k(2)}{F_k(1)} = frac{g(2)^2}{g(1)^2} $$
is the square of a rational number, and thus the square of a positive integer $m$. But $1 + 2^k = m^2$ means $(m-1)(m+1) = 2^k$, so both $m-1$ and $m+1$ are powers of $2$. The only powers of $2$ that differ by $2$ are $2$ and $2^2$, which gives us the case $k=3$.
answered Jan 17 at 3:32
Robert IsraelRobert Israel
324k23214468
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add a comment |
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$begingroup$
There is no other $k$.
Faulhaber's formula for $F_k(x)$ has rational coefficients and leading term $1/(k+1)$. If $F_k(x)$ is the square of a polynomial, we have
$$ F_k(x) = frac{g(x)^2}{k+1}$$
where $g(x)^2$ is monic with rational coefficients. Now $g(x)$, being the square root of a monic polynomial with rational coefficients, has rational coefficients (in fact the square root of a monic polynomial over the rationals can be written as a Laurent series over the rationals; it is a polynomial if the terms in negative powers of $x$ are all $0$).
In particular,
$$ 1 + 2^k = frac{F_k(2)}{F_k(1)} = frac{g(2)^2}{g(1)^2} $$
is the square of a rational number, and thus the square of a positive integer $m$. But $1 + 2^k = m^2$ means $(m-1)(m+1) = 2^k$, so both $m-1$ and $m+1$ are powers of $2$. The only powers of $2$ that differ by $2$ are $2$ and $2^2$, which gives us the case $k=3$.
answered Jan 17 at 3:32
Robert IsraelRobert Israel
324k23214468
$endgroup$
add a comment |
$begingroup$
There is no other $k$.
Faulhaber's formula for $F_k(x)$ has rational coefficients and leading term $1/(k+1)$. If $F_k(x)$ is the square of a polynomial, we have
$$ F_k(x) = frac{g(x)^2}{k+1}$$
where $g(x)^2$ is monic with rational coefficients. Now $g(x)$, being the square root of a monic polynomial with rational coefficients, has rational coefficients (in fact the square root of a monic polynomial over the rationals can be written as a Laurent series over the rationals; it is a polynomial if the terms in negative powers of $x$ are all $0$).
In particular,
$$ 1 + 2^k = frac{F_k(2)}{F_k(1)} = frac{g(2)^2}{g(1)^2} $$
is the square of a rational number, and thus the square of a positive integer $m$. But $1 + 2^k = m^2$ means $(m-1)(m+1) = 2^k$, so both $m-1$ and $m+1$ are powers of $2$. The only powers of $2$ that differ by $2$ are $2$ and $2^2$, which gives us the case $k=3$.
answered Jan 17 at 3:32
Robert IsraelRobert Israel
324k23214468
$endgroup$
add a comment |
$begingroup$
There is no other $k$.
Faulhaber's formula for $F_k(x)$ has rational coefficients and leading term $1/(k+1)$. If $F_k(x)$ is the square of a polynomial, we have
$$ F_k(x) = frac{g(x)^2}{k+1}$$
where $g(x)^2$ is monic with rational coefficients. Now $g(x)$, being the square root of a monic polynomial with rational coefficients, has rational coefficients (in fact the square root of a monic polynomial over the rationals can be written as a Laurent series over the rationals; it is a polynomial if the terms in negative powers of $x$ are all $0$).
In particular,
$$ 1 + 2^k = frac{F_k(2)}{F_k(1)} = frac{g(2)^2}{g(1)^2} $$
is the square of a rational number, and thus the square of a positive integer $m$. But $1 + 2^k = m^2$ means $(m-1)(m+1) = 2^k$, so both $m-1$ and $m+1$ are powers of $2$. The only powers of $2$ that differ by $2$ are $2$ and $2^2$, which gives us the case $k=3$.
answered Jan 17 at 3:32
Robert IsraelRobert Israel
324k23214468
$endgroup$
There is no other $k$.
Faulhaber's formula for $F_k(x)$ has rational coefficients and leading term $1/(k+1)$. If $F_k(x)$ is the square of a polynomial, we have
$$ F_k(x) = frac{g(x)^2}{k+1}$$
where $g(x)^2$ is monic with rational coefficients. Now $g(x)$, being the square root of a monic polynomial with rational coefficients, has rational coefficients (in fact the square root of a monic polynomial over the rationals can be written as a Laurent series over the rationals; it is a polynomial if the terms in negative powers of $x$ are all $0$).
In particular,
$$ 1 + 2^k = frac{F_k(2)}{F_k(1)} = frac{g(2)^2}{g(1)^2} $$
is the square of a rational number, and thus the square of a positive integer $m$. But $1 + 2^k = m^2$ means $(m-1)(m+1) = 2^k$, so both $m-1$ and $m+1$ are powers of $2$. The only powers of $2$ that differ by $2$ are $2$ and $2^2$, which gives us the case $k=3$.
answered Jan 17 at 3:32
Robert IsraelRobert Israel
324k23214468
edited Jan 17 at 12:20
answered Jan 17 at 3:32
Robert IsraelRobert Israel
324k23214468
answered Jan 17 at 3:32
Robert IsraelRobert Israel
324k23214468
answered Jan 17 at 3:32
Robert IsraelRobert Israel
324k23214468
324k23214468
add a comment |
add a comment |
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$begingroup$
I do not think so but it would nice since it would be an extension of Nicomachus's theorem.
$endgroup$
– Claude Leibovici
Jan 17 at 3:20
1
$begingroup$
Bernoulli gave a way to obtain general algebraic expressions for what you call $P_k(n)$ for any $k$.
$endgroup$
– Keith Backman
Jan 17 at 3:40