Find all primes $p$ such that $left( frac{600}{p} right )= -1$
0
$begingroup$
How to solve this one?
I understand that 600 is $2^2 * 3 * 5^2$, but I don't know how to continue to solve it.
Thank you up front.
number-theory prime-numbers
asked Jan 16 at 9:31
Igor Igor
63
$endgroup$
|
show 10 more comments
0
$begingroup$
How to solve this one?
I understand that 600 is $2^2 * 3 * 5^2$, but I don't know how to continue to solve it.
Thank you up front.
number-theory prime-numbers
asked Jan 16 at 9:31
Igor Igor
63
$endgroup$
2
$begingroup$
What do you mean by $()$? Also, $600=2^3times3times5^2$.
$endgroup$
– Dan Uznanski
Jan 16 at 9:33
$begingroup$
math.stackexchange.com/questions/1998653/…
$endgroup$
– lab bhattacharjee
Jan 16 at 9:33
$begingroup$
@labbhattacharjee How should I break the 600, in this case, i'm not sure what to do with the powers on prime numbers
$endgroup$
– Igor
Jan 16 at 9:35
$begingroup$
I assume you mean the Legendre symbol. Observe $left(frac{2^2}{p}right)=1$ for all $p>3$, same for $5^2$.
$endgroup$
– Wojowu
Jan 16 at 10:01
$begingroup$
@Wojowu Do you mean $left( frac{2^2}{p} right )$ $=1$ or $=-1$.
$endgroup$
– Igor
Jan 16 at 10:06
|
show 10 more comments
0
0
0
$begingroup$
How to solve this one?
I understand that 600 is $2^2 * 3 * 5^2$, but I don't know how to continue to solve it.
Thank you up front.
number-theory prime-numbers
asked Jan 16 at 9:31
Igor Igor
63
$endgroup$
How to solve this one?
I understand that 600 is $2^2 * 3 * 5^2$, but I don't know how to continue to solve it.
Thank you up front.
number-theory prime-numbers
number-theory prime-numbers
asked Jan 16 at 9:31
Igor Igor
63
asked Jan 16 at 9:31
Igor Igor
63
asked Jan 16 at 9:31
Igor Igor
63
asked Jan 16 at 9:31
Igor Igor
63
asked Jan 16 at 9:31
Igor Igor
63
63
2
$begingroup$
What do you mean by $()$? Also, $600=2^3times3times5^2$.
$endgroup$
– Dan Uznanski
Jan 16 at 9:33
$begingroup$
math.stackexchange.com/questions/1998653/…
$endgroup$
– lab bhattacharjee
Jan 16 at 9:33
$begingroup$
@labbhattacharjee How should I break the 600, in this case, i'm not sure what to do with the powers on prime numbers
$endgroup$
– Igor
Jan 16 at 9:35
$begingroup$
I assume you mean the Legendre symbol. Observe $left(frac{2^2}{p}right)=1$ for all $p>3$, same for $5^2$.
$endgroup$
– Wojowu
Jan 16 at 10:01
$begingroup$
@Wojowu Do you mean $left( frac{2^2}{p} right )$ $=1$ or $=-1$.
$endgroup$
– Igor
Jan 16 at 10:06
|
show 10 more comments
2
$begingroup$
What do you mean by $()$? Also, $600=2^3times3times5^2$.
$endgroup$
– Dan Uznanski
Jan 16 at 9:33
$begingroup$
math.stackexchange.com/questions/1998653/…
$endgroup$
– lab bhattacharjee
Jan 16 at 9:33
$begingroup$
@labbhattacharjee How should I break the 600, in this case, i'm not sure what to do with the powers on prime numbers
$endgroup$
– Igor
Jan 16 at 9:35
$begingroup$
I assume you mean the Legendre symbol. Observe $left(frac{2^2}{p}right)=1$ for all $p>3$, same for $5^2$.
$endgroup$
– Wojowu
Jan 16 at 10:01
$begingroup$
@Wojowu Do you mean $left( frac{2^2}{p} right )$ $=1$ or $=-1$.
$endgroup$
– Igor
Jan 16 at 10:06
2
2
$begingroup$
What do you mean by $()$? Also, $600=2^3times3times5^2$.
$endgroup$
– Dan Uznanski
Jan 16 at 9:33
$begingroup$
What do you mean by $()$? Also, $600=2^3times3times5^2$.
$endgroup$
– Dan Uznanski
Jan 16 at 9:33
$begingroup$
math.stackexchange.com/questions/1998653/…
$endgroup$
– lab bhattacharjee
Jan 16 at 9:33
$begingroup$
math.stackexchange.com/questions/1998653/…
$endgroup$
– lab bhattacharjee
Jan 16 at 9:33
$begingroup$
@labbhattacharjee How should I break the 600, in this case, i'm not sure what to do with the powers on prime numbers
$endgroup$
– Igor
Jan 16 at 9:35
$begingroup$
@labbhattacharjee How should I break the 600, in this case, i'm not sure what to do with the powers on prime numbers
$endgroup$
– Igor
Jan 16 at 9:35
$begingroup$
I assume you mean the Legendre symbol. Observe $left(frac{2^2}{p}right)=1$ for all $p>3$, same for $5^2$.
$endgroup$
– Wojowu
Jan 16 at 10:01
$begingroup$
I assume you mean the Legendre symbol. Observe $left(frac{2^2}{p}right)=1$ for all $p>3$, same for $5^2$.
$endgroup$
– Wojowu
Jan 16 at 10:01
$begingroup$
@Wojowu Do you mean $left( frac{2^2}{p} right )$ $=1$ or $=-1$.
$endgroup$
– Igor
Jan 16 at 10:06
$begingroup$
@Wojowu Do you mean $left( frac{2^2}{p} right )$ $=1$ or $=-1$.
$endgroup$
– Igor
Jan 16 at 10:06
|
show 10 more comments
0
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2
$begingroup$
What do you mean by $()$? Also, $600=2^3times3times5^2$.
$endgroup$
– Dan Uznanski
Jan 16 at 9:33
$begingroup$
math.stackexchange.com/questions/1998653/…
$endgroup$
– lab bhattacharjee
Jan 16 at 9:33
$begingroup$
@labbhattacharjee How should I break the 600, in this case, i'm not sure what to do with the powers on prime numbers
$endgroup$
– Igor
Jan 16 at 9:35
$begingroup$
I assume you mean the Legendre symbol. Observe $left(frac{2^2}{p}right)=1$ for all $p>3$, same for $5^2$.
$endgroup$
– Wojowu
Jan 16 at 10:01
$begingroup$
@Wojowu Do you mean $left( frac{2^2}{p} right )$ $=1$ or $=-1$.
$endgroup$
– Igor
Jan 16 at 10:06