Determine the quadratic residues $QR_{85}$.
$begingroup$
Define the set of quadratic residues as $QR_N = {x^2 bmod N: x in mathbb{Z}_N^*}$.
I'm asked to compute $QR_{85}$ without any further knowledge.
I know that there are $phi(85) = phi(5) phi(17) = 4 cdot 16 = 64$ units to test.
What observation can allow me to reduce the computation?
Here is a python script to compute them:
for i in range(0,85):
if(i % 5 != 0 and i % 17 != 0):
print(i," ",i*i % 85)
As pointed out in the comments $mathbb{Z}^{*}_{85} cong mathbb{Z}_4 times mathbb{Z}_{16}$. The squares in $mathbb{Z}_4 times mathbb{Z}_{16}$ are the tuples $(l,r)$ with $l in {0,2}$ and $r in {0,2,4,6,8,10,12,14}$.
The mapping to pass from $mathbb{Z}_4 times mathbb{Z}_{16}$ to $mathbb{Z}^{*}_5 times mathbb{Z}^{*}_{17}$ is trivial and the mapping from $mathbb{Z}^{*}_5 times mathbb{Z}^{*}_{17}$ to $mathbb{Z}^{*}_{85}$ is $h^{-1}(u,v) = -34 u + 35 v bmod 85$
I wonder if there is an easier way to do this.
abstract-algebra quadratic-residues
edited Dec 28 '18 at 5:28
Shaun
9,570113684
asked Dec 27 '18 at 17:38
JavierJavier
2,07521234
$endgroup$
add a comment |
$begingroup$
Define the set of quadratic residues as $QR_N = {x^2 bmod N: x in mathbb{Z}_N^*}$.
I'm asked to compute $QR_{85}$ without any further knowledge.
I know that there are $phi(85) = phi(5) phi(17) = 4 cdot 16 = 64$ units to test.
What observation can allow me to reduce the computation?
Here is a python script to compute them:
for i in range(0,85):
if(i % 5 != 0 and i % 17 != 0):
print(i," ",i*i % 85)
As pointed out in the comments $mathbb{Z}^{*}_{85} cong mathbb{Z}_4 times mathbb{Z}_{16}$. The squares in $mathbb{Z}_4 times mathbb{Z}_{16}$ are the tuples $(l,r)$ with $l in {0,2}$ and $r in {0,2,4,6,8,10,12,14}$.
The mapping to pass from $mathbb{Z}_4 times mathbb{Z}_{16}$ to $mathbb{Z}^{*}_5 times mathbb{Z}^{*}_{17}$ is trivial and the mapping from $mathbb{Z}^{*}_5 times mathbb{Z}^{*}_{17}$ to $mathbb{Z}^{*}_{85}$ is $h^{-1}(u,v) = -34 u + 35 v bmod 85$
I wonder if there is an easier way to do this.
abstract-algebra quadratic-residues
edited Dec 28 '18 at 5:28
Shaun
9,570113684
asked Dec 27 '18 at 17:38
JavierJavier
2,07521234
$endgroup$
1
$begingroup$
By the Chinese remainder theorem this unit group is isomorphic to $Bbb{Z}/4Bbb{Z}timesBbb{Z}/16Bbb{Z}$, and the quadratic residues correspond to the multiples of $2$.
$endgroup$
– Servaes
Dec 27 '18 at 18:34
add a comment |
$begingroup$
Define the set of quadratic residues as $QR_N = {x^2 bmod N: x in mathbb{Z}_N^*}$.
I'm asked to compute $QR_{85}$ without any further knowledge.
I know that there are $phi(85) = phi(5) phi(17) = 4 cdot 16 = 64$ units to test.
What observation can allow me to reduce the computation?
Here is a python script to compute them:
for i in range(0,85):
if(i % 5 != 0 and i % 17 != 0):
print(i," ",i*i % 85)
As pointed out in the comments $mathbb{Z}^{*}_{85} cong mathbb{Z}_4 times mathbb{Z}_{16}$. The squares in $mathbb{Z}_4 times mathbb{Z}_{16}$ are the tuples $(l,r)$ with $l in {0,2}$ and $r in {0,2,4,6,8,10,12,14}$.
The mapping to pass from $mathbb{Z}_4 times mathbb{Z}_{16}$ to $mathbb{Z}^{*}_5 times mathbb{Z}^{*}_{17}$ is trivial and the mapping from $mathbb{Z}^{*}_5 times mathbb{Z}^{*}_{17}$ to $mathbb{Z}^{*}_{85}$ is $h^{-1}(u,v) = -34 u + 35 v bmod 85$
I wonder if there is an easier way to do this.
abstract-algebra quadratic-residues
edited Dec 28 '18 at 5:28
Shaun
9,570113684
asked Dec 27 '18 at 17:38
JavierJavier
2,07521234
$endgroup$
Define the set of quadratic residues as $QR_N = {x^2 bmod N: x in mathbb{Z}_N^*}$.
I'm asked to compute $QR_{85}$ without any further knowledge.
I know that there are $phi(85) = phi(5) phi(17) = 4 cdot 16 = 64$ units to test.
What observation can allow me to reduce the computation?
Here is a python script to compute them:
for i in range(0,85):
if(i % 5 != 0 and i % 17 != 0):
print(i," ",i*i % 85)
As pointed out in the comments $mathbb{Z}^{*}_{85} cong mathbb{Z}_4 times mathbb{Z}_{16}$. The squares in $mathbb{Z}_4 times mathbb{Z}_{16}$ are the tuples $(l,r)$ with $l in {0,2}$ and $r in {0,2,4,6,8,10,12,14}$.
The mapping to pass from $mathbb{Z}_4 times mathbb{Z}_{16}$ to $mathbb{Z}^{*}_5 times mathbb{Z}^{*}_{17}$ is trivial and the mapping from $mathbb{Z}^{*}_5 times mathbb{Z}^{*}_{17}$ to $mathbb{Z}^{*}_{85}$ is $h^{-1}(u,v) = -34 u + 35 v bmod 85$
I wonder if there is an easier way to do this.
abstract-algebra quadratic-residues
abstract-algebra quadratic-residues
edited Dec 28 '18 at 5:28
Shaun
9,570113684
asked Dec 27 '18 at 17:38
JavierJavier
2,07521234
edited Dec 28 '18 at 5:28
Shaun
9,570113684
asked Dec 27 '18 at 17:38
JavierJavier
2,07521234
edited Dec 28 '18 at 5:28
Shaun
9,570113684
edited Dec 28 '18 at 5:28
Shaun
9,570113684
edited Dec 28 '18 at 5:28
Shaun
9,570113684
9,570113684
asked Dec 27 '18 at 17:38
JavierJavier
2,07521234
asked Dec 27 '18 at 17:38
JavierJavier
2,07521234
asked Dec 27 '18 at 17:38
JavierJavier
2,07521234
2,07521234
1
$begingroup$
By the Chinese remainder theorem this unit group is isomorphic to $Bbb{Z}/4Bbb{Z}timesBbb{Z}/16Bbb{Z}$, and the quadratic residues correspond to the multiples of $2$.
$endgroup$
– Servaes
Dec 27 '18 at 18:34
add a comment |
1
$begingroup$
By the Chinese remainder theorem this unit group is isomorphic to $Bbb{Z}/4Bbb{Z}timesBbb{Z}/16Bbb{Z}$, and the quadratic residues correspond to the multiples of $2$.
$endgroup$
– Servaes
Dec 27 '18 at 18:34
1
1
$begingroup$
By the Chinese remainder theorem this unit group is isomorphic to $Bbb{Z}/4Bbb{Z}timesBbb{Z}/16Bbb{Z}$, and the quadratic residues correspond to the multiples of $2$.
$endgroup$
– Servaes
Dec 27 '18 at 18:34
$begingroup$
By the Chinese remainder theorem this unit group is isomorphic to $Bbb{Z}/4Bbb{Z}timesBbb{Z}/16Bbb{Z}$, and the quadratic residues correspond to the multiples of $2$.
$endgroup$
– Servaes
Dec 27 '18 at 18:34
add a comment |
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$begingroup$
By the Chinese remainder theorem this unit group is isomorphic to $Bbb{Z}/4Bbb{Z}timesBbb{Z}/16Bbb{Z}$, and the quadratic residues correspond to the multiples of $2$.
$endgroup$
– Servaes
Dec 27 '18 at 18:34