Mixing
Find the pairs $(m,k)$ solving the diophantine equation $m^2=7k+9$
$begingroup$
Solve
$$m^2=7k+9$$
over the integers
First i rearrange got $m^2-9=7k$
And $(m^2-9)/7=k$
So first $m^2-9$ must be divisible by $7$
So suppose $m=7n , 7n+1 ,...,7n+5$ But it doesn't work
....
number-theory diophantine-equations integers
edited Jan 11 at 14:48
Peter
47.4k1039129
asked Jan 11 at 12:53
HeartHeart
26717
$endgroup$
add a comment |
$begingroup$
Solve
$$m^2=7k+9$$
over the integers
First i rearrange got $m^2-9=7k$
And $(m^2-9)/7=k$
So first $m^2-9$ must be divisible by $7$
So suppose $m=7n , 7n+1 ,...,7n+5$ But it doesn't work
....
number-theory diophantine-equations integers
edited Jan 11 at 14:48
Peter
47.4k1039129
asked Jan 11 at 12:53
HeartHeart
26717
$endgroup$
1
$begingroup$
So what you're looking at is $$ m^2 equiv 2 {mod 7} $$ I would suggest plugging in some values for $m$ and seeing what happens. $m=1, 2, 3, 4, 5, 6, 7 ldots$. If you find a solution $m_0$,then $m_0 pm 7$ is also a solution. Therefore, you only need to look at the first $7$ integers.
$endgroup$
– Matti P.
Jan 11 at 12:57
2
$begingroup$
Hint: $m^2-9=(m-3)(m+3)$, so $7mid m^2-9iff7mid m-3lor7mid m+3$, since $7$ is prime.
$endgroup$
– Barry Cipra
Jan 11 at 14:57
$begingroup$
If $(m_0,k_0)$ is a solution, then $(m_0+7, k_0+2m_0+7)$ is also a solution $$m_0^2=7k_0+9 iff m_0^2+14m_0+49=7k_0+9+14m_0+49=7(k_0+2m_0+7)+9 iff\ (m_0+7)^2=7(k_0+2m_0+7)+9$$ and start with $(3,0)$.
$endgroup$
– rtybase
Jan 11 at 17:49
add a comment |
$begingroup$
Solve
$$m^2=7k+9$$
over the integers
First i rearrange got $m^2-9=7k$
And $(m^2-9)/7=k$
So first $m^2-9$ must be divisible by $7$
So suppose $m=7n , 7n+1 ,...,7n+5$ But it doesn't work
....
number-theory diophantine-equations integers
edited Jan 11 at 14:48
Peter
47.4k1039129
asked Jan 11 at 12:53
HeartHeart
26717
$endgroup$
Solve
$$m^2=7k+9$$
over the integers
First i rearrange got $m^2-9=7k$
And $(m^2-9)/7=k$
So first $m^2-9$ must be divisible by $7$
So suppose $m=7n , 7n+1 ,...,7n+5$ But it doesn't work
....
number-theory diophantine-equations integers
number-theory diophantine-equations integers
edited Jan 11 at 14:48
Peter
47.4k1039129
asked Jan 11 at 12:53
HeartHeart
26717
edited Jan 11 at 14:48
Peter
47.4k1039129
asked Jan 11 at 12:53
HeartHeart
26717
edited Jan 11 at 14:48
Peter
47.4k1039129
edited Jan 11 at 14:48
Peter
47.4k1039129
edited Jan 11 at 14:48
Peter
47.4k1039129
47.4k1039129
asked Jan 11 at 12:53
HeartHeart
26717
asked Jan 11 at 12:53
HeartHeart
26717
asked Jan 11 at 12:53
HeartHeart
26717
26717
1
$begingroup$
So what you're looking at is $$ m^2 equiv 2 {mod 7} $$ I would suggest plugging in some values for $m$ and seeing what happens. $m=1, 2, 3, 4, 5, 6, 7 ldots$. If you find a solution $m_0$,then $m_0 pm 7$ is also a solution. Therefore, you only need to look at the first $7$ integers.
$endgroup$
– Matti P.
Jan 11 at 12:57
2
$begingroup$
Hint: $m^2-9=(m-3)(m+3)$, so $7mid m^2-9iff7mid m-3lor7mid m+3$, since $7$ is prime.
$endgroup$
– Barry Cipra
Jan 11 at 14:57
$begingroup$
If $(m_0,k_0)$ is a solution, then $(m_0+7, k_0+2m_0+7)$ is also a solution $$m_0^2=7k_0+9 iff m_0^2+14m_0+49=7k_0+9+14m_0+49=7(k_0+2m_0+7)+9 iff\ (m_0+7)^2=7(k_0+2m_0+7)+9$$ and start with $(3,0)$.
$endgroup$
– rtybase
Jan 11 at 17:49
add a comment |
1
$begingroup$
So what you're looking at is $$ m^2 equiv 2 {mod 7} $$ I would suggest plugging in some values for $m$ and seeing what happens. $m=1, 2, 3, 4, 5, 6, 7 ldots$. If you find a solution $m_0$,then $m_0 pm 7$ is also a solution. Therefore, you only need to look at the first $7$ integers.
$endgroup$
– Matti P.
Jan 11 at 12:57
2
$begingroup$
Hint: $m^2-9=(m-3)(m+3)$, so $7mid m^2-9iff7mid m-3lor7mid m+3$, since $7$ is prime.
$endgroup$
– Barry Cipra
Jan 11 at 14:57
$begingroup$
If $(m_0,k_0)$ is a solution, then $(m_0+7, k_0+2m_0+7)$ is also a solution $$m_0^2=7k_0+9 iff m_0^2+14m_0+49=7k_0+9+14m_0+49=7(k_0+2m_0+7)+9 iff\ (m_0+7)^2=7(k_0+2m_0+7)+9$$ and start with $(3,0)$.
$endgroup$
– rtybase
Jan 11 at 17:49
1
1
$begingroup$
So what you're looking at is $$ m^2 equiv 2 {mod 7} $$ I would suggest plugging in some values for $m$ and seeing what happens. $m=1, 2, 3, 4, 5, 6, 7 ldots$. If you find a solution $m_0$,then $m_0 pm 7$ is also a solution. Therefore, you only need to look at the first $7$ integers.
$endgroup$
– Matti P.
Jan 11 at 12:57
$begingroup$
So what you're looking at is $$ m^2 equiv 2 {mod 7} $$ I would suggest plugging in some values for $m$ and seeing what happens. $m=1, 2, 3, 4, 5, 6, 7 ldots$. If you find a solution $m_0$,then $m_0 pm 7$ is also a solution. Therefore, you only need to look at the first $7$ integers.
$endgroup$
– Matti P.
Jan 11 at 12:57
2
2
$begingroup$
Hint: $m^2-9=(m-3)(m+3)$, so $7mid m^2-9iff7mid m-3lor7mid m+3$, since $7$ is prime.
$endgroup$
– Barry Cipra
Jan 11 at 14:57
$begingroup$
Hint: $m^2-9=(m-3)(m+3)$, so $7mid m^2-9iff7mid m-3lor7mid m+3$, since $7$ is prime.
$endgroup$
– Barry Cipra
Jan 11 at 14:57
$begingroup$
If $(m_0,k_0)$ is a solution, then $(m_0+7, k_0+2m_0+7)$ is also a solution $$m_0^2=7k_0+9 iff m_0^2+14m_0+49=7k_0+9+14m_0+49=7(k_0+2m_0+7)+9 iff\ (m_0+7)^2=7(k_0+2m_0+7)+9$$ and start with $(3,0)$.
$endgroup$
– rtybase
Jan 11 at 17:49
$begingroup$
If $(m_0,k_0)$ is a solution, then $(m_0+7, k_0+2m_0+7)$ is also a solution $$m_0^2=7k_0+9 iff m_0^2+14m_0+49=7k_0+9+14m_0+49=7(k_0+2m_0+7)+9 iff\ (m_0+7)^2=7(k_0+2m_0+7)+9$$ and start with $(3,0)$.
$endgroup$
– rtybase
Jan 11 at 17:49
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
$m^2=7k+9 Rightarrow m^2=2 ;(mod 7)$.
The only numbers from 1 to 7 satisfying the equation are 3 and 4 ($3^2=9=2; 4^2=16=2$. Then, the solution is of the form:
$min{3+7n, 4+7n, n in mathbb{N}}$; and $k=frac{m^2-9}{7}$
For instance, if $m=81=4+77$, then $k=936$, and
$$81^2=7 cdot 936 + 9$$
answered Jan 11 at 17:22
pendermathpendermath
58412
$endgroup$
add a comment |
$begingroup$
Above equation $m^2=7k+9$ has solution's,
$m=7w-4$
$k=(7w-1)(w-1)$, where, 'w' is any integer
For $w=2$ we get, $(m.k)=(10,13)$
$(10)^2=7(13)+9$
edited Jan 12 at 23:18
amWhy
1
answered Jan 11 at 17:26
SamSam
11
$endgroup$
$begingroup$
What about (3,0)?
$endgroup$
– pendermath
Jan 11 at 17:27
$begingroup$
It is included $w=1$
$endgroup$
– Mike
Jan 11 at 18:31
$begingroup$
Sorry, I meant (4,1)
$endgroup$
– pendermath
Jan 11 at 21:04
$begingroup$
For w=0, you will get (m,k)=(4,1)
$endgroup$
– Sam
Jan 11 at 22:10
$begingroup$
If $w=0$, then $m=-4$, not 4, right?
$endgroup$
– pendermath
Jan 12 at 22:37
add a comment |
$begingroup$
As you said $7$ must divide $m^2-9=(m-3)(m+3)$. Since $7$ is prime, it must divide either $m-3$ or $m+3$.
Therefore, you have two sets of solutions:
Case 1: $7|m-3$ then
$$m-3=7l \
m=7l+3 \
7k+9=m^2=49l^2+42l+9 \
k=7l^2+6l\
(k,m)= ( 7l^2+6l, 7l+3)$$
Case 2: $7|m+3$ then
$$m+3=7l \
m=7l-3 \
7k+9=m^2=49l^2-42l+9 \
k=7l^2+6l\
(k,m)= ( 7l^2-6l, 7l-3)$$
answered Jan 12 at 23:45
N. S.N. S.
103k6112208
$endgroup$
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
$begingroup$
$m^2=7k+9 Rightarrow m^2=2 ;(mod 7)$.
The only numbers from 1 to 7 satisfying the equation are 3 and 4 ($3^2=9=2; 4^2=16=2$. Then, the solution is of the form:
$min{3+7n, 4+7n, n in mathbb{N}}$; and $k=frac{m^2-9}{7}$
For instance, if $m=81=4+77$, then $k=936$, and
$$81^2=7 cdot 936 + 9$$
answered Jan 11 at 17:22
pendermathpendermath
58412
$endgroup$
add a comment |
$begingroup$
$m^2=7k+9 Rightarrow m^2=2 ;(mod 7)$.
The only numbers from 1 to 7 satisfying the equation are 3 and 4 ($3^2=9=2; 4^2=16=2$. Then, the solution is of the form:
$min{3+7n, 4+7n, n in mathbb{N}}$; and $k=frac{m^2-9}{7}$
For instance, if $m=81=4+77$, then $k=936$, and
$$81^2=7 cdot 936 + 9$$
answered Jan 11 at 17:22
pendermathpendermath
58412
$endgroup$
add a comment |
$begingroup$
$m^2=7k+9 Rightarrow m^2=2 ;(mod 7)$.
The only numbers from 1 to 7 satisfying the equation are 3 and 4 ($3^2=9=2; 4^2=16=2$. Then, the solution is of the form:
$min{3+7n, 4+7n, n in mathbb{N}}$; and $k=frac{m^2-9}{7}$
For instance, if $m=81=4+77$, then $k=936$, and
$$81^2=7 cdot 936 + 9$$
answered Jan 11 at 17:22
pendermathpendermath
58412
$endgroup$
$m^2=7k+9 Rightarrow m^2=2 ;(mod 7)$.
The only numbers from 1 to 7 satisfying the equation are 3 and 4 ($3^2=9=2; 4^2=16=2$. Then, the solution is of the form:
$min{3+7n, 4+7n, n in mathbb{N}}$; and $k=frac{m^2-9}{7}$
For instance, if $m=81=4+77$, then $k=936$, and
$$81^2=7 cdot 936 + 9$$
answered Jan 11 at 17:22
pendermathpendermath
58412
answered Jan 11 at 17:22
pendermathpendermath
58412
answered Jan 11 at 17:22
pendermathpendermath
58412
answered Jan 11 at 17:22
pendermathpendermath
58412
58412
add a comment |
add a comment |
$begingroup$
Above equation $m^2=7k+9$ has solution's,
$m=7w-4$
$k=(7w-1)(w-1)$, where, 'w' is any integer
For $w=2$ we get, $(m.k)=(10,13)$
$(10)^2=7(13)+9$
edited Jan 12 at 23:18
amWhy
1
answered Jan 11 at 17:26
SamSam
11
$endgroup$
$begingroup$
What about (3,0)?
$endgroup$
– pendermath
Jan 11 at 17:27
$begingroup$
It is included $w=1$
$endgroup$
– Mike
Jan 11 at 18:31
$begingroup$
Sorry, I meant (4,1)
$endgroup$
– pendermath
Jan 11 at 21:04
$begingroup$
For w=0, you will get (m,k)=(4,1)
$endgroup$
– Sam
Jan 11 at 22:10
$begingroup$
If $w=0$, then $m=-4$, not 4, right?
$endgroup$
– pendermath
Jan 12 at 22:37
add a comment |
$begingroup$
Above equation $m^2=7k+9$ has solution's,
$m=7w-4$
$k=(7w-1)(w-1)$, where, 'w' is any integer
For $w=2$ we get, $(m.k)=(10,13)$
$(10)^2=7(13)+9$
edited Jan 12 at 23:18
amWhy
1
answered Jan 11 at 17:26
SamSam
11
$endgroup$
$begingroup$
What about (3,0)?
$endgroup$
– pendermath
Jan 11 at 17:27
$begingroup$
It is included $w=1$
$endgroup$
– Mike
Jan 11 at 18:31
$begingroup$
Sorry, I meant (4,1)
$endgroup$
– pendermath
Jan 11 at 21:04
$begingroup$
For w=0, you will get (m,k)=(4,1)
$endgroup$
– Sam
Jan 11 at 22:10
$begingroup$
If $w=0$, then $m=-4$, not 4, right?
$endgroup$
– pendermath
Jan 12 at 22:37
add a comment |
$begingroup$
Above equation $m^2=7k+9$ has solution's,
$m=7w-4$
$k=(7w-1)(w-1)$, where, 'w' is any integer
For $w=2$ we get, $(m.k)=(10,13)$
$(10)^2=7(13)+9$
edited Jan 12 at 23:18
amWhy
1
answered Jan 11 at 17:26
SamSam
11
$endgroup$
Above equation $m^2=7k+9$ has solution's,
$m=7w-4$
$k=(7w-1)(w-1)$, where, 'w' is any integer
For $w=2$ we get, $(m.k)=(10,13)$
$(10)^2=7(13)+9$
edited Jan 12 at 23:18
amWhy
1
answered Jan 11 at 17:26
SamSam
11
edited Jan 12 at 23:18
amWhy
1
edited Jan 12 at 23:18
amWhy
1
edited Jan 12 at 23:18
amWhy
1
1
answered Jan 11 at 17:26
SamSam
11
answered Jan 11 at 17:26
SamSam
11
answered Jan 11 at 17:26
SamSam
11
11
$begingroup$
What about (3,0)?
$endgroup$
– pendermath
Jan 11 at 17:27
$begingroup$
It is included $w=1$
$endgroup$
– Mike
Jan 11 at 18:31
$begingroup$
Sorry, I meant (4,1)
$endgroup$
– pendermath
Jan 11 at 21:04
$begingroup$
For w=0, you will get (m,k)=(4,1)
$endgroup$
– Sam
Jan 11 at 22:10
$begingroup$
If $w=0$, then $m=-4$, not 4, right?
$endgroup$
– pendermath
Jan 12 at 22:37
add a comment |
$begingroup$
What about (3,0)?
$endgroup$
– pendermath
Jan 11 at 17:27
$begingroup$
It is included $w=1$
$endgroup$
– Mike
Jan 11 at 18:31
$begingroup$
Sorry, I meant (4,1)
$endgroup$
– pendermath
Jan 11 at 21:04
$begingroup$
For w=0, you will get (m,k)=(4,1)
$endgroup$
– Sam
Jan 11 at 22:10
$begingroup$
If $w=0$, then $m=-4$, not 4, right?
$endgroup$
– pendermath
Jan 12 at 22:37
$begingroup$
What about (3,0)?
$endgroup$
– pendermath
Jan 11 at 17:27
$begingroup$
What about (3,0)?
$endgroup$
– pendermath
Jan 11 at 17:27
$begingroup$
It is included $w=1$
$endgroup$
– Mike
Jan 11 at 18:31
$begingroup$
It is included $w=1$
$endgroup$
– Mike
Jan 11 at 18:31
$begingroup$
Sorry, I meant (4,1)
$endgroup$
– pendermath
Jan 11 at 21:04
$begingroup$
Sorry, I meant (4,1)
$endgroup$
– pendermath
Jan 11 at 21:04
$begingroup$
For w=0, you will get (m,k)=(4,1)
$endgroup$
– Sam
Jan 11 at 22:10
$begingroup$
For w=0, you will get (m,k)=(4,1)
$endgroup$
– Sam
Jan 11 at 22:10
$begingroup$
If $w=0$, then $m=-4$, not 4, right?
$endgroup$
– pendermath
Jan 12 at 22:37
$begingroup$
If $w=0$, then $m=-4$, not 4, right?
$endgroup$
– pendermath
Jan 12 at 22:37
add a comment |
$begingroup$
As you said $7$ must divide $m^2-9=(m-3)(m+3)$. Since $7$ is prime, it must divide either $m-3$ or $m+3$.
Therefore, you have two sets of solutions:
Case 1: $7|m-3$ then
$$m-3=7l \
m=7l+3 \
7k+9=m^2=49l^2+42l+9 \
k=7l^2+6l\
(k,m)= ( 7l^2+6l, 7l+3)$$
Case 2: $7|m+3$ then
$$m+3=7l \
m=7l-3 \
7k+9=m^2=49l^2-42l+9 \
k=7l^2+6l\
(k,m)= ( 7l^2-6l, 7l-3)$$
answered Jan 12 at 23:45
N. S.N. S.
103k6112208
$endgroup$
add a comment |
$begingroup$
As you said $7$ must divide $m^2-9=(m-3)(m+3)$. Since $7$ is prime, it must divide either $m-3$ or $m+3$.
Therefore, you have two sets of solutions:
Case 1: $7|m-3$ then
$$m-3=7l \
m=7l+3 \
7k+9=m^2=49l^2+42l+9 \
k=7l^2+6l\
(k,m)= ( 7l^2+6l, 7l+3)$$
Case 2: $7|m+3$ then
$$m+3=7l \
m=7l-3 \
7k+9=m^2=49l^2-42l+9 \
k=7l^2+6l\
(k,m)= ( 7l^2-6l, 7l-3)$$
answered Jan 12 at 23:45
N. S.N. S.
103k6112208
$endgroup$
add a comment |
$begingroup$
As you said $7$ must divide $m^2-9=(m-3)(m+3)$. Since $7$ is prime, it must divide either $m-3$ or $m+3$.
Therefore, you have two sets of solutions:
Case 1: $7|m-3$ then
$$m-3=7l \
m=7l+3 \
7k+9=m^2=49l^2+42l+9 \
k=7l^2+6l\
(k,m)= ( 7l^2+6l, 7l+3)$$
Case 2: $7|m+3$ then
$$m+3=7l \
m=7l-3 \
7k+9=m^2=49l^2-42l+9 \
k=7l^2+6l\
(k,m)= ( 7l^2-6l, 7l-3)$$
answered Jan 12 at 23:45
N. S.N. S.
103k6112208
$endgroup$
As you said $7$ must divide $m^2-9=(m-3)(m+3)$. Since $7$ is prime, it must divide either $m-3$ or $m+3$.
Therefore, you have two sets of solutions:
Case 1: $7|m-3$ then
$$m-3=7l \
m=7l+3 \
7k+9=m^2=49l^2+42l+9 \
k=7l^2+6l\
(k,m)= ( 7l^2+6l, 7l+3)$$
Case 2: $7|m+3$ then
$$m+3=7l \
m=7l-3 \
7k+9=m^2=49l^2-42l+9 \
k=7l^2+6l\
(k,m)= ( 7l^2-6l, 7l-3)$$
answered Jan 12 at 23:45
N. S.N. S.
103k6112208
answered Jan 12 at 23:45
N. S.N. S.
103k6112208
answered Jan 12 at 23:45
N. S.N. S.
103k6112208
answered Jan 12 at 23:45
N. S.N. S.
103k6112208
103k6112208
add a comment |
add a comment |
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1
$begingroup$
So what you're looking at is $$ m^2 equiv 2 {mod 7} $$ I would suggest plugging in some values for $m$ and seeing what happens. $m=1, 2, 3, 4, 5, 6, 7 ldots$. If you find a solution $m_0$,then $m_0 pm 7$ is also a solution. Therefore, you only need to look at the first $7$ integers.
$endgroup$
– Matti P.
Jan 11 at 12:57
2
$begingroup$
Hint: $m^2-9=(m-3)(m+3)$, so $7mid m^2-9iff7mid m-3lor7mid m+3$, since $7$ is prime.
$endgroup$
– Barry Cipra
Jan 11 at 14:57
$begingroup$
If $(m_0,k_0)$ is a solution, then $(m_0+7, k_0+2m_0+7)$ is also a solution $$m_0^2=7k_0+9 iff m_0^2+14m_0+49=7k_0+9+14m_0+49=7(k_0+2m_0+7)+9 iff\ (m_0+7)^2=7(k_0+2m_0+7)+9$$ and start with $(3,0)$.
$endgroup$
– rtybase
Jan 11 at 17:49