How to prove $sum_d tau^3(d)=left(sum_dtau(d)right)^2$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Help with $sum_dmid nτ(d)^2=sum_d mid nτ(d)^3$Proving the identity $sum_k=1^n k^3 = big(sum_k=1^n kbig)^2$ without inductionProof of $sum_d tau^3(d)=left(sum_dtau(d)right)^2$ (not standard proof)How prove this $sum_n(d(t))^3=left(sum_nd(t)right)^2$Ramanujan's Tau function, an arithmetic propertyProve that $tau(n) leq 2sqrtn$How prove this $sum_n(d(t))^3=left(sum_nd(t)right)^2$Help with $sum_dmid nτ(d)^2=sum_d mid nτ(d)^3$Prove that $sum limits_d(n/d)sigma(d) = sum limits_ddtau(d)$Prove or disprove: $ sum_b vee d = x tau(b) tau(d) = tau(x)^3$For a given integer $n$, how many primes $p_1,p_2 leq n$ such that $tau(p_1-1)=tau(p_2-1)$Proof of sum of positive divisors of $n$ (probably repeated question somewhere in the stack)Has $sigmaleft(sigma_0(n)^4right)=n$ infinitely many solutions?$phi(n)^sigma(n)^tau(n)=n^2$ find all natural numbers $n$ such that the equality is true
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How to prove $sum_d tau^3(d)=left(sum_dtau(d)right)^2$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Help with $sum_dmid nτ(d)^2=sum_d mid nτ(d)^3$Proving the identity $sum_k=1^n k^3 = big(sum_k=1^n kbig)^2$ without inductionProof of $sum_d tau^3(d)=left(sum_dtau(d)right)^2$ (not standard proof)How prove this $sum_t(d(t))^3=left(sum_td(t)right)^2$Ramanujan's Tau function, an arithmetic propertyProve that $tau(n) leq 2sqrtn$How prove this $sum_t(d(t))^3=left(sum_td(t)right)^2$Help with $sum_dmid nτ(d)^2=sum_d mid nτ(d)^3$Prove that $sum limits_d(n/d)sigma(d) = sum limits_ddtau(d)$Prove or disprove: $ sum_b vee d = x tau(b) tau(d) = tau(x)^3$For a given integer $n$, how many primes $p_1,p_2 leq n$ such that $tau(p_1-1)=tau(p_2-1)$Proof of sum of positive divisors of $n$ (probably repeated question somewhere in the stack)Has $sigmaleft(sigma_0(n)^4right)=n$ infinitely many solutions?$phi(n)^sigma(n)^tau(n)=n^2$ find all natural numbers $n$ such that the equality is true
$begingroup$
For every positive integer $d$, we let $tauleft(dright)$ be the number of positive divisors of $d$.
Prove that
beginalign
sum_d tau^3(d)
= left(sum_d tau (d)right)^2
endalign
for each positive integer $n$, where the sums range over all positive divisors $d$ of $n$.
Now I only know that both sides are multiplicative arithmetic functions in $n$. Could you tell me what I need to do next?
elementary-number-theory analytic-number-theory arithmetic-functions
edited Apr 1 at 18:05
darij grinberg
11.5k33168
asked Nov 2 '11 at 23:54
VladimirVladimir
1,11311231
$endgroup$
add a comment |
$begingroup$
For every positive integer $d$, we let $tauleft(dright)$ be the number of positive divisors of $d$.
Prove that
beginalign
sum_d tau^3(d)
= left(sum_d tau (d)right)^2
endalign
for each positive integer $n$, where the sums range over all positive divisors $d$ of $n$.
Now I only know that both sides are multiplicative arithmetic functions in $n$. Could you tell me what I need to do next?
elementary-number-theory analytic-number-theory arithmetic-functions
edited Apr 1 at 18:05
darij grinberg
11.5k33168
asked Nov 2 '11 at 23:54
VladimirVladimir
1,11311231
$endgroup$
2
$begingroup$
If you have proved that both sides are multiplicative functions, it suffices to prove that the statement holds for powers of primes. In this case, it is straight forward to compute the value of each side. It may be useful to use the fact that $sum_i=1^n i^3=(sum_i=1^n i)^2$.
$endgroup$
– Aaron
Nov 3 '11 at 0:06
add a comment |
$begingroup$
For every positive integer $d$, we let $tauleft(dright)$ be the number of positive divisors of $d$.
Prove that
beginalign
sum_d tau^3(d)
= left(sum_d tau (d)right)^2
endalign
for each positive integer $n$, where the sums range over all positive divisors $d$ of $n$.
Now I only know that both sides are multiplicative arithmetic functions in $n$. Could you tell me what I need to do next?
elementary-number-theory analytic-number-theory arithmetic-functions
edited Apr 1 at 18:05
darij grinberg
11.5k33168
asked Nov 2 '11 at 23:54
VladimirVladimir
1,11311231
$endgroup$
For every positive integer $d$, we let $tauleft(dright)$ be the number of positive divisors of $d$.
Prove that
beginalign
sum_d tau^3(d)
= left(sum_d tau (d)right)^2
endalign
for each positive integer $n$, where the sums range over all positive divisors $d$ of $n$.
Now I only know that both sides are multiplicative arithmetic functions in $n$. Could you tell me what I need to do next?
elementary-number-theory analytic-number-theory arithmetic-functions
elementary-number-theory analytic-number-theory arithmetic-functions
edited Apr 1 at 18:05
darij grinberg
11.5k33168
asked Nov 2 '11 at 23:54
VladimirVladimir
1,11311231
edited Apr 1 at 18:05
darij grinberg
11.5k33168
asked Nov 2 '11 at 23:54
VladimirVladimir
1,11311231
edited Apr 1 at 18:05
darij grinberg
11.5k33168
edited Apr 1 at 18:05
darij grinberg
11.5k33168
edited Apr 1 at 18:05
darij grinberg
11.5k33168
11.5k33168
asked Nov 2 '11 at 23:54
VladimirVladimir
1,11311231
asked Nov 2 '11 at 23:54
VladimirVladimir
1,11311231
asked Nov 2 '11 at 23:54
VladimirVladimir
1,11311231
1,11311231
2
$begingroup$
If you have proved that both sides are multiplicative functions, it suffices to prove that the statement holds for powers of primes. In this case, it is straight forward to compute the value of each side. It may be useful to use the fact that $sum_i=1^n i^3=(sum_i=1^n i)^2$.
$endgroup$
– Aaron
Nov 3 '11 at 0:06
add a comment |
2
$begingroup$
If you have proved that both sides are multiplicative functions, it suffices to prove that the statement holds for powers of primes. In this case, it is straight forward to compute the value of each side. It may be useful to use the fact that $sum_i=1^n i^3=(sum_i=1^n i)^2$.
$endgroup$
– Aaron
Nov 3 '11 at 0:06
2
2
$begingroup$
If you have proved that both sides are multiplicative functions, it suffices to prove that the statement holds for powers of primes. In this case, it is straight forward to compute the value of each side. It may be useful to use the fact that $sum_i=1^n i^3=(sum_i=1^n i)^2$.
$endgroup$
– Aaron
Nov 3 '11 at 0:06
$begingroup$
If you have proved that both sides are multiplicative functions, it suffices to prove that the statement holds for powers of primes. In this case, it is straight forward to compute the value of each side. It may be useful to use the fact that $sum_i=1^n i^3=(sum_i=1^n i)^2$.
$endgroup$
– Aaron
Nov 3 '11 at 0:06
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
If you've shown that both the LHS and the RHS are multiplicative functions, then you must now show it's true for arbitrary prime powers $n=p^r$. In doing so, use $tau(n)=sigma_1(p^r)=r+1$ and this. It then follows for all composite numbers by prime factorizing both sides through the multiplication.
edited Apr 13 '17 at 12:21
Community♦
1
answered Nov 3 '11 at 0:05
anonanon
72.6k5112217
$endgroup$
add a comment |
$begingroup$
Recall that if $f$ is a multiplicative function, and
$$g(n)=sum_dfleft(fracndright),$$
then $g$ is a multiplicative function.
From this you can deduce that both sides of your identity are multiplicative functions. Thus to verify the identity, all you need to do is to verify it for $n=p^k$, where $p$ is prime.
For $n=p^k$, the left-hand side is $1^3+2^3+cdots +k^3+(k+1)^3$ (since $p^i$ has $i+1$ positive divisors for each $i geq 0$). The right-hand side is $(1+2+cdots+k+(k+1))^2$. The fact that the two are equal is a probably familiar fact. If it isn't, it can be proved by induction.
edited Apr 7 at 17:50
darij grinberg
11.5k33168
answered Nov 3 '11 at 0:05
André NicolasAndré Nicolas
455k36432822
$endgroup$
$begingroup$
Where did you get that "Recall that if $f$ is a multiplicative function..." formula? Is that a known identity?
$endgroup$
– TeaFor2
Mar 21 '18 at 15:59
add a comment |
$begingroup$
To get this result you need to show
(a) it is true for prime powers, which you can show with $$ sum_1^n i^3 = fracn^2(n+1)^24 = left(sum_1^n iright)^2$$
(b) both the left and right sides are multiplicative, which you say you know
answered Nov 3 '11 at 0:12
HenryHenry
101k482170
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you've shown that both the LHS and the RHS are multiplicative functions, then you must now show it's true for arbitrary prime powers $n=p^r$. In doing so, use $tau(n)=sigma_1(p^r)=r+1$ and this. It then follows for all composite numbers by prime factorizing both sides through the multiplication.
edited Apr 13 '17 at 12:21
Community♦
1
answered Nov 3 '11 at 0:05
anonanon
72.6k5112217
$endgroup$
add a comment |
$begingroup$
If you've shown that both the LHS and the RHS are multiplicative functions, then you must now show it's true for arbitrary prime powers $n=p^r$. In doing so, use $tau(n)=sigma_1(p^r)=r+1$ and this. It then follows for all composite numbers by prime factorizing both sides through the multiplication.
edited Apr 13 '17 at 12:21
Community♦
1
answered Nov 3 '11 at 0:05
anonanon
72.6k5112217
$endgroup$
add a comment |
$begingroup$
If you've shown that both the LHS and the RHS are multiplicative functions, then you must now show it's true for arbitrary prime powers $n=p^r$. In doing so, use $tau(n)=sigma_1(p^r)=r+1$ and this. It then follows for all composite numbers by prime factorizing both sides through the multiplication.
edited Apr 13 '17 at 12:21
Community♦
1
answered Nov 3 '11 at 0:05
anonanon
72.6k5112217
$endgroup$
If you've shown that both the LHS and the RHS are multiplicative functions, then you must now show it's true for arbitrary prime powers $n=p^r$. In doing so, use $tau(n)=sigma_1(p^r)=r+1$ and this. It then follows for all composite numbers by prime factorizing both sides through the multiplication.
edited Apr 13 '17 at 12:21
Community♦
1
answered Nov 3 '11 at 0:05
anonanon
72.6k5112217
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
1
answered Nov 3 '11 at 0:05
anonanon
72.6k5112217
answered Nov 3 '11 at 0:05
anonanon
72.6k5112217
answered Nov 3 '11 at 0:05
anonanon
72.6k5112217
72.6k5112217
add a comment |
add a comment |
$begingroup$
Recall that if $f$ is a multiplicative function, and
$$g(n)=sum_dfleft(fracndright),$$
then $g$ is a multiplicative function.
From this you can deduce that both sides of your identity are multiplicative functions. Thus to verify the identity, all you need to do is to verify it for $n=p^k$, where $p$ is prime.
For $n=p^k$, the left-hand side is $1^3+2^3+cdots +k^3+(k+1)^3$ (since $p^i$ has $i+1$ positive divisors for each $i geq 0$). The right-hand side is $(1+2+cdots+k+(k+1))^2$. The fact that the two are equal is a probably familiar fact. If it isn't, it can be proved by induction.
edited Apr 7 at 17:50
darij grinberg
11.5k33168
answered Nov 3 '11 at 0:05
André NicolasAndré Nicolas
455k36432822
$endgroup$
$begingroup$
Where did you get that "Recall that if $f$ is a multiplicative function..." formula? Is that a known identity?
$endgroup$
– TeaFor2
Mar 21 '18 at 15:59
add a comment |
$begingroup$
Recall that if $f$ is a multiplicative function, and
$$g(n)=sum_dfleft(fracndright),$$
then $g$ is a multiplicative function.
From this you can deduce that both sides of your identity are multiplicative functions. Thus to verify the identity, all you need to do is to verify it for $n=p^k$, where $p$ is prime.
For $n=p^k$, the left-hand side is $1^3+2^3+cdots +k^3+(k+1)^3$ (since $p^i$ has $i+1$ positive divisors for each $i geq 0$). The right-hand side is $(1+2+cdots+k+(k+1))^2$. The fact that the two are equal is a probably familiar fact. If it isn't, it can be proved by induction.
edited Apr 7 at 17:50
darij grinberg
11.5k33168
answered Nov 3 '11 at 0:05
André NicolasAndré Nicolas
455k36432822
$endgroup$
$begingroup$
Where did you get that "Recall that if $f$ is a multiplicative function..." formula? Is that a known identity?
$endgroup$
– TeaFor2
Mar 21 '18 at 15:59
add a comment |
$begingroup$
Recall that if $f$ is a multiplicative function, and
$$g(n)=sum_dfleft(fracndright),$$
then $g$ is a multiplicative function.
From this you can deduce that both sides of your identity are multiplicative functions. Thus to verify the identity, all you need to do is to verify it for $n=p^k$, where $p$ is prime.
For $n=p^k$, the left-hand side is $1^3+2^3+cdots +k^3+(k+1)^3$ (since $p^i$ has $i+1$ positive divisors for each $i geq 0$). The right-hand side is $(1+2+cdots+k+(k+1))^2$. The fact that the two are equal is a probably familiar fact. If it isn't, it can be proved by induction.
edited Apr 7 at 17:50
darij grinberg
11.5k33168
answered Nov 3 '11 at 0:05
André NicolasAndré Nicolas
455k36432822
$endgroup$
Recall that if $f$ is a multiplicative function, and
$$g(n)=sum_dfleft(fracndright),$$
then $g$ is a multiplicative function.
From this you can deduce that both sides of your identity are multiplicative functions. Thus to verify the identity, all you need to do is to verify it for $n=p^k$, where $p$ is prime.
For $n=p^k$, the left-hand side is $1^3+2^3+cdots +k^3+(k+1)^3$ (since $p^i$ has $i+1$ positive divisors for each $i geq 0$). The right-hand side is $(1+2+cdots+k+(k+1))^2$. The fact that the two are equal is a probably familiar fact. If it isn't, it can be proved by induction.
edited Apr 7 at 17:50
darij grinberg
11.5k33168
answered Nov 3 '11 at 0:05
André NicolasAndré Nicolas
455k36432822
edited Apr 7 at 17:50
darij grinberg
11.5k33168
edited Apr 7 at 17:50
darij grinberg
11.5k33168
edited Apr 7 at 17:50
darij grinberg
11.5k33168
11.5k33168
answered Nov 3 '11 at 0:05
André NicolasAndré Nicolas
455k36432822
answered Nov 3 '11 at 0:05
André NicolasAndré Nicolas
455k36432822
answered Nov 3 '11 at 0:05
André NicolasAndré Nicolas
455k36432822
455k36432822
$begingroup$
Where did you get that "Recall that if $f$ is a multiplicative function..." formula? Is that a known identity?
$endgroup$
– TeaFor2
Mar 21 '18 at 15:59
add a comment |
$begingroup$
Where did you get that "Recall that if $f$ is a multiplicative function..." formula? Is that a known identity?
$endgroup$
– TeaFor2
Mar 21 '18 at 15:59
$begingroup$
Where did you get that "Recall that if $f$ is a multiplicative function..." formula? Is that a known identity?
$endgroup$
– TeaFor2
Mar 21 '18 at 15:59
$begingroup$
Where did you get that "Recall that if $f$ is a multiplicative function..." formula? Is that a known identity?
$endgroup$
– TeaFor2
Mar 21 '18 at 15:59
add a comment |
$begingroup$
To get this result you need to show
(a) it is true for prime powers, which you can show with $$ sum_1^n i^3 = fracn^2(n+1)^24 = left(sum_1^n iright)^2$$
(b) both the left and right sides are multiplicative, which you say you know
answered Nov 3 '11 at 0:12
HenryHenry
101k482170
$endgroup$
add a comment |
$begingroup$
To get this result you need to show
(a) it is true for prime powers, which you can show with $$ sum_1^n i^3 = fracn^2(n+1)^24 = left(sum_1^n iright)^2$$
(b) both the left and right sides are multiplicative, which you say you know
answered Nov 3 '11 at 0:12
HenryHenry
101k482170
$endgroup$
add a comment |
$begingroup$
To get this result you need to show
(a) it is true for prime powers, which you can show with $$ sum_1^n i^3 = fracn^2(n+1)^24 = left(sum_1^n iright)^2$$
(b) both the left and right sides are multiplicative, which you say you know
answered Nov 3 '11 at 0:12
HenryHenry
101k482170
$endgroup$
To get this result you need to show
(a) it is true for prime powers, which you can show with $$ sum_1^n i^3 = fracn^2(n+1)^24 = left(sum_1^n iright)^2$$
(b) both the left and right sides are multiplicative, which you say you know
answered Nov 3 '11 at 0:12
HenryHenry
101k482170
answered Nov 3 '11 at 0:12
HenryHenry
101k482170
answered Nov 3 '11 at 0:12
HenryHenry
101k482170
answered Nov 3 '11 at 0:12
HenryHenry
101k482170
101k482170
add a comment |
add a comment |
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$begingroup$
If you have proved that both sides are multiplicative functions, it suffices to prove that the statement holds for powers of primes. In this case, it is straight forward to compute the value of each side. It may be useful to use the fact that $sum_i=1^n i^3=(sum_i=1^n i)^2$.
$endgroup$
– Aaron
Nov 3 '11 at 0:06