Find all integers $x,y$ whose squares sum up to $c$ mod 5
$begingroup$
Find all integers $x,y$ such that $x^2 + y^2 equiv c pmod{5}$.
I managed to solve by trying one-by-one, but I guess there is some other way to solve this?
modular-arithmetic
edited Jan 18 at 9:09
Yuval Filmus
49k472148
asked Jan 18 at 8:52
Katrine NicolaisenKatrine Nicolaisen
1
$endgroup$
add a comment |
$begingroup$
Find all integers $x,y$ such that $x^2 + y^2 equiv c pmod{5}$.
I managed to solve by trying one-by-one, but I guess there is some other way to solve this?
modular-arithmetic
edited Jan 18 at 9:09
Yuval Filmus
49k472148
asked Jan 18 at 8:52
Katrine NicolaisenKatrine Nicolaisen
1
$endgroup$
1
$begingroup$
Please rewrite the equation in the post itself. What do you want it to be equivalent to in mod 5 ? And I think you can just consider what values $n^2$ gets in general in mod 5.
$endgroup$
– Matti P.
Jan 18 at 8:53
add a comment |
$begingroup$
Find all integers $x,y$ such that $x^2 + y^2 equiv c pmod{5}$.
I managed to solve by trying one-by-one, but I guess there is some other way to solve this?
modular-arithmetic
edited Jan 18 at 9:09
Yuval Filmus
49k472148
asked Jan 18 at 8:52
Katrine NicolaisenKatrine Nicolaisen
1
$endgroup$
Find all integers $x,y$ such that $x^2 + y^2 equiv c pmod{5}$.
I managed to solve by trying one-by-one, but I guess there is some other way to solve this?
modular-arithmetic
modular-arithmetic
edited Jan 18 at 9:09
Yuval Filmus
49k472148
asked Jan 18 at 8:52
Katrine NicolaisenKatrine Nicolaisen
1
edited Jan 18 at 9:09
Yuval Filmus
49k472148
asked Jan 18 at 8:52
Katrine NicolaisenKatrine Nicolaisen
1
edited Jan 18 at 9:09
Yuval Filmus
49k472148
edited Jan 18 at 9:09
Yuval Filmus
49k472148
edited Jan 18 at 9:09
Yuval Filmus
49k472148
49k472148
asked Jan 18 at 8:52
Katrine NicolaisenKatrine Nicolaisen
1
asked Jan 18 at 8:52
Katrine NicolaisenKatrine Nicolaisen
1
asked Jan 18 at 8:52
Katrine NicolaisenKatrine Nicolaisen
1
1
1
$begingroup$
Please rewrite the equation in the post itself. What do you want it to be equivalent to in mod 5 ? And I think you can just consider what values $n^2$ gets in general in mod 5.
$endgroup$
– Matti P.
Jan 18 at 8:53
add a comment |
1
$begingroup$
Please rewrite the equation in the post itself. What do you want it to be equivalent to in mod 5 ? And I think you can just consider what values $n^2$ gets in general in mod 5.
$endgroup$
– Matti P.
Jan 18 at 8:53
1
1
$begingroup$
Please rewrite the equation in the post itself. What do you want it to be equivalent to in mod 5 ? And I think you can just consider what values $n^2$ gets in general in mod 5.
$endgroup$
– Matti P.
Jan 18 at 8:53
$begingroup$
Please rewrite the equation in the post itself. What do you want it to be equivalent to in mod 5 ? And I think you can just consider what values $n^2$ gets in general in mod 5.
$endgroup$
– Matti P.
Jan 18 at 8:53
add a comment |
1 Answer
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$begingroup$
Now we have that all the squares the integers numbers with last digit $1,4,6,9$ will have last digit $ equiv 1 mod 5$. Same argument if the last digit is $2,3,7,8$ is $ equiv 4 mod 5$. And we have $ equiv 0 mod 5$ if the last digit is $0,5$.
Thus if
$c=0$ then $x=5k$, $y=5j$ for $k,j in mathbb{Z}$
$ x=5k+1$ or $x=5k+4$, $y=5j+2$ or $y=5j+3$ for $k,j in mathbb{Z}$
$c=1$ then $x=5k$, $y=5j+1$ or $y=5j+4$ for $k,j in mathbb{Z}$
$c=2$ then $x=5k+1$ or $x=5k+4$, $y=5j+1$ or $y=5j+4$ for $k,j in mathbb{Z}$
$c=3$ then $x=5k+2$ or $x=5k+3$, $y=5j+2$ or $y=5j+3$ for $k,j in mathbb{Z}$
$c=4$ then $x=5k$, $y=5j+2$ or $y=5j+3$ for $k,j in mathbb{Z}$
Since this expression in symmetric in $x$ and $y$ you can also swap $x$ and $y$
answered Jan 18 at 9:20
user289143user289143
1,069313
$endgroup$
$begingroup$
Thank you very mouch:-)
$endgroup$
– Katrine Nicolaisen
Jan 21 at 22:49
add a comment |
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$begingroup$
Now we have that all the squares the integers numbers with last digit $1,4,6,9$ will have last digit $ equiv 1 mod 5$. Same argument if the last digit is $2,3,7,8$ is $ equiv 4 mod 5$. And we have $ equiv 0 mod 5$ if the last digit is $0,5$.
Thus if
$c=0$ then $x=5k$, $y=5j$ for $k,j in mathbb{Z}$
$ x=5k+1$ or $x=5k+4$, $y=5j+2$ or $y=5j+3$ for $k,j in mathbb{Z}$
$c=1$ then $x=5k$, $y=5j+1$ or $y=5j+4$ for $k,j in mathbb{Z}$
$c=2$ then $x=5k+1$ or $x=5k+4$, $y=5j+1$ or $y=5j+4$ for $k,j in mathbb{Z}$
$c=3$ then $x=5k+2$ or $x=5k+3$, $y=5j+2$ or $y=5j+3$ for $k,j in mathbb{Z}$
$c=4$ then $x=5k$, $y=5j+2$ or $y=5j+3$ for $k,j in mathbb{Z}$
Since this expression in symmetric in $x$ and $y$ you can also swap $x$ and $y$
answered Jan 18 at 9:20
user289143user289143
1,069313
$endgroup$
$begingroup$
Thank you very mouch:-)
$endgroup$
– Katrine Nicolaisen
Jan 21 at 22:49
add a comment |
$begingroup$
Now we have that all the squares the integers numbers with last digit $1,4,6,9$ will have last digit $ equiv 1 mod 5$. Same argument if the last digit is $2,3,7,8$ is $ equiv 4 mod 5$. And we have $ equiv 0 mod 5$ if the last digit is $0,5$.
Thus if
$c=0$ then $x=5k$, $y=5j$ for $k,j in mathbb{Z}$
$ x=5k+1$ or $x=5k+4$, $y=5j+2$ or $y=5j+3$ for $k,j in mathbb{Z}$
$c=1$ then $x=5k$, $y=5j+1$ or $y=5j+4$ for $k,j in mathbb{Z}$
$c=2$ then $x=5k+1$ or $x=5k+4$, $y=5j+1$ or $y=5j+4$ for $k,j in mathbb{Z}$
$c=3$ then $x=5k+2$ or $x=5k+3$, $y=5j+2$ or $y=5j+3$ for $k,j in mathbb{Z}$
$c=4$ then $x=5k$, $y=5j+2$ or $y=5j+3$ for $k,j in mathbb{Z}$
Since this expression in symmetric in $x$ and $y$ you can also swap $x$ and $y$
answered Jan 18 at 9:20
user289143user289143
1,069313
$endgroup$
$begingroup$
Thank you very mouch:-)
$endgroup$
– Katrine Nicolaisen
Jan 21 at 22:49
add a comment |
$begingroup$
Now we have that all the squares the integers numbers with last digit $1,4,6,9$ will have last digit $ equiv 1 mod 5$. Same argument if the last digit is $2,3,7,8$ is $ equiv 4 mod 5$. And we have $ equiv 0 mod 5$ if the last digit is $0,5$.
Thus if
$c=0$ then $x=5k$, $y=5j$ for $k,j in mathbb{Z}$
$ x=5k+1$ or $x=5k+4$, $y=5j+2$ or $y=5j+3$ for $k,j in mathbb{Z}$
$c=1$ then $x=5k$, $y=5j+1$ or $y=5j+4$ for $k,j in mathbb{Z}$
$c=2$ then $x=5k+1$ or $x=5k+4$, $y=5j+1$ or $y=5j+4$ for $k,j in mathbb{Z}$
$c=3$ then $x=5k+2$ or $x=5k+3$, $y=5j+2$ or $y=5j+3$ for $k,j in mathbb{Z}$
$c=4$ then $x=5k$, $y=5j+2$ or $y=5j+3$ for $k,j in mathbb{Z}$
Since this expression in symmetric in $x$ and $y$ you can also swap $x$ and $y$
answered Jan 18 at 9:20
user289143user289143
1,069313
$endgroup$
Now we have that all the squares the integers numbers with last digit $1,4,6,9$ will have last digit $ equiv 1 mod 5$. Same argument if the last digit is $2,3,7,8$ is $ equiv 4 mod 5$. And we have $ equiv 0 mod 5$ if the last digit is $0,5$.
Thus if
$c=0$ then $x=5k$, $y=5j$ for $k,j in mathbb{Z}$
$ x=5k+1$ or $x=5k+4$, $y=5j+2$ or $y=5j+3$ for $k,j in mathbb{Z}$
$c=1$ then $x=5k$, $y=5j+1$ or $y=5j+4$ for $k,j in mathbb{Z}$
$c=2$ then $x=5k+1$ or $x=5k+4$, $y=5j+1$ or $y=5j+4$ for $k,j in mathbb{Z}$
$c=3$ then $x=5k+2$ or $x=5k+3$, $y=5j+2$ or $y=5j+3$ for $k,j in mathbb{Z}$
$c=4$ then $x=5k$, $y=5j+2$ or $y=5j+3$ for $k,j in mathbb{Z}$
Since this expression in symmetric in $x$ and $y$ you can also swap $x$ and $y$
answered Jan 18 at 9:20
user289143user289143
1,069313
answered Jan 18 at 9:20
user289143user289143
1,069313
answered Jan 18 at 9:20
user289143user289143
1,069313
answered Jan 18 at 9:20
user289143user289143
1,069313
1,069313
$begingroup$
Thank you very mouch:-)
$endgroup$
– Katrine Nicolaisen
Jan 21 at 22:49
add a comment |
$begingroup$
Thank you very mouch:-)
$endgroup$
– Katrine Nicolaisen
Jan 21 at 22:49
$begingroup$
Thank you very mouch:-)
$endgroup$
– Katrine Nicolaisen
Jan 21 at 22:49
$begingroup$
Thank you very mouch:-)
$endgroup$
– Katrine Nicolaisen
Jan 21 at 22:49
add a comment |
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$begingroup$
Please rewrite the equation in the post itself. What do you want it to be equivalent to in mod 5 ? And I think you can just consider what values $n^2$ gets in general in mod 5.
$endgroup$
– Matti P.
Jan 18 at 8:53