How to solve these two simultaneous “divisibilities” : $n+1mid m^2+1$ and $m+1mid n^2+1$
$begingroup$
Is it possible to find all integers $m>0$ and $n>0$ such that $n+1mid m^2+1$ and $m+1,|,n^2+1$ ?
I succeed to prove there is an infinite number of solutions, but I cannot progress anymore.
Thanks !
number-theory elementary-number-theory divisibility diophantine-equations congruences
asked Feb 25 '16 at 7:49
uvdoseuvdose
259514
$endgroup$
|
show 5 more comments
$begingroup$
Is it possible to find all integers $m>0$ and $n>0$ such that $n+1mid m^2+1$ and $m+1,|,n^2+1$ ?
I succeed to prove there is an infinite number of solutions, but I cannot progress anymore.
Thanks !
number-theory elementary-number-theory divisibility diophantine-equations congruences
asked Feb 25 '16 at 7:49
uvdoseuvdose
259514
$endgroup$
12
$begingroup$
Could you write down your proof for the infinite number of solutions? It may help everyone.
$endgroup$
– user37238
Feb 25 '16 at 8:22
3
$begingroup$
If you understand French, there is a long discussion (partly numerical, partly theoretical) here
$endgroup$
– Giovanni Resta
Feb 25 '16 at 8:28
2
$begingroup$
Yes, I understand French: in fact, I participated in this discussion. One of the participants (serge17) managed to find a method to build "big" solutions $(m,n)$ : Using his method, we can prove that there are infinitely many solutions.
$endgroup$
– uvdose
Feb 25 '16 at 8:37
4
$begingroup$
@individ : for $N=1,2,3,4$, the solutions $(m,n)$ are : $(33,217)$ , $(22641,961753)$ , $(263568049,55479822393)$ , $(45074835574129,41942086060150713)$.
$endgroup$
– uvdose
Feb 29 '16 at 14:25
5
$begingroup$
If (n+1) | m^2 + 1 and (m+ 1) | n^2 + 1 then (n + 1) | ( m + 1) ^2 - 2m and ((m+1) ^ 2) | n^4 + 2 (m^2) + 1 therefore ( n + 1) | (n^4 + 2( n^2) + 1)/ h - 2m , for some integer h . So ( n+ 1) | (n^4 + 2( n^2) + 1 - (2m h)) , so (n+1) | (4n^2 - (2 m h)) and ( n+1) | 2( m h +2) . 2| m iff 2| m , so if 2| n then ( n+1) | (m h +2) and similarly ( m+1) | ( n k+ 2) for some k an element of integers. I don't know if this helps....
$endgroup$
– 201044
Mar 15 '16 at 23:39
|
show 5 more comments
$begingroup$
Is it possible to find all integers $m>0$ and $n>0$ such that $n+1mid m^2+1$ and $m+1,|,n^2+1$ ?
I succeed to prove there is an infinite number of solutions, but I cannot progress anymore.
Thanks !
number-theory elementary-number-theory divisibility diophantine-equations congruences
asked Feb 25 '16 at 7:49
uvdoseuvdose
259514
$endgroup$
Is it possible to find all integers $m>0$ and $n>0$ such that $n+1mid m^2+1$ and $m+1,|,n^2+1$ ?
I succeed to prove there is an infinite number of solutions, but I cannot progress anymore.
Thanks !
number-theory elementary-number-theory divisibility diophantine-equations congruences
number-theory elementary-number-theory divisibility diophantine-equations congruences
asked Feb 25 '16 at 7:49
uvdoseuvdose
259514
asked Feb 25 '16 at 7:49
uvdoseuvdose
259514
edited Dec 30 '16 at 16:56
uvdose
asked Feb 25 '16 at 7:49
uvdoseuvdose
259514
asked Feb 25 '16 at 7:49
uvdoseuvdose
259514
asked Feb 25 '16 at 7:49
uvdoseuvdose
259514
259514
12
$begingroup$
Could you write down your proof for the infinite number of solutions? It may help everyone.
$endgroup$
– user37238
Feb 25 '16 at 8:22
3
$begingroup$
If you understand French, there is a long discussion (partly numerical, partly theoretical) here
$endgroup$
– Giovanni Resta
Feb 25 '16 at 8:28
2
$begingroup$
Yes, I understand French: in fact, I participated in this discussion. One of the participants (serge17) managed to find a method to build "big" solutions $(m,n)$ : Using his method, we can prove that there are infinitely many solutions.
$endgroup$
– uvdose
Feb 25 '16 at 8:37
4
$begingroup$
@individ : for $N=1,2,3,4$, the solutions $(m,n)$ are : $(33,217)$ , $(22641,961753)$ , $(263568049,55479822393)$ , $(45074835574129,41942086060150713)$.
$endgroup$
– uvdose
Feb 29 '16 at 14:25
5
$begingroup$
If (n+1) | m^2 + 1 and (m+ 1) | n^2 + 1 then (n + 1) | ( m + 1) ^2 - 2m and ((m+1) ^ 2) | n^4 + 2 (m^2) + 1 therefore ( n + 1) | (n^4 + 2( n^2) + 1)/ h - 2m , for some integer h . So ( n+ 1) | (n^4 + 2( n^2) + 1 - (2m h)) , so (n+1) | (4n^2 - (2 m h)) and ( n+1) | 2( m h +2) . 2| m iff 2| m , so if 2| n then ( n+1) | (m h +2) and similarly ( m+1) | ( n k+ 2) for some k an element of integers. I don't know if this helps....
$endgroup$
– 201044
Mar 15 '16 at 23:39
|
show 5 more comments
12
$begingroup$
Could you write down your proof for the infinite number of solutions? It may help everyone.
$endgroup$
– user37238
Feb 25 '16 at 8:22
3
$begingroup$
If you understand French, there is a long discussion (partly numerical, partly theoretical) here
$endgroup$
– Giovanni Resta
Feb 25 '16 at 8:28
2
$begingroup$
Yes, I understand French: in fact, I participated in this discussion. One of the participants (serge17) managed to find a method to build "big" solutions $(m,n)$ : Using his method, we can prove that there are infinitely many solutions.
$endgroup$
– uvdose
Feb 25 '16 at 8:37
4
$begingroup$
@individ : for $N=1,2,3,4$, the solutions $(m,n)$ are : $(33,217)$ , $(22641,961753)$ , $(263568049,55479822393)$ , $(45074835574129,41942086060150713)$.
$endgroup$
– uvdose
Feb 29 '16 at 14:25
5
$begingroup$
If (n+1) | m^2 + 1 and (m+ 1) | n^2 + 1 then (n + 1) | ( m + 1) ^2 - 2m and ((m+1) ^ 2) | n^4 + 2 (m^2) + 1 therefore ( n + 1) | (n^4 + 2( n^2) + 1)/ h - 2m , for some integer h . So ( n+ 1) | (n^4 + 2( n^2) + 1 - (2m h)) , so (n+1) | (4n^2 - (2 m h)) and ( n+1) | 2( m h +2) . 2| m iff 2| m , so if 2| n then ( n+1) | (m h +2) and similarly ( m+1) | ( n k+ 2) for some k an element of integers. I don't know if this helps....
$endgroup$
– 201044
Mar 15 '16 at 23:39
12
12
$begingroup$
Could you write down your proof for the infinite number of solutions? It may help everyone.
$endgroup$
– user37238
Feb 25 '16 at 8:22
$begingroup$
Could you write down your proof for the infinite number of solutions? It may help everyone.
$endgroup$
– user37238
Feb 25 '16 at 8:22
3
3
$begingroup$
If you understand French, there is a long discussion (partly numerical, partly theoretical) here
$endgroup$
– Giovanni Resta
Feb 25 '16 at 8:28
$begingroup$
If you understand French, there is a long discussion (partly numerical, partly theoretical) here
$endgroup$
– Giovanni Resta
Feb 25 '16 at 8:28
2
2
$begingroup$
Yes, I understand French: in fact, I participated in this discussion. One of the participants (serge17) managed to find a method to build "big" solutions $(m,n)$ : Using his method, we can prove that there are infinitely many solutions.
$endgroup$
– uvdose
Feb 25 '16 at 8:37
$begingroup$
Yes, I understand French: in fact, I participated in this discussion. One of the participants (serge17) managed to find a method to build "big" solutions $(m,n)$ : Using his method, we can prove that there are infinitely many solutions.
$endgroup$
– uvdose
Feb 25 '16 at 8:37
4
4
$begingroup$
@individ : for $N=1,2,3,4$, the solutions $(m,n)$ are : $(33,217)$ , $(22641,961753)$ , $(263568049,55479822393)$ , $(45074835574129,41942086060150713)$.
$endgroup$
– uvdose
Feb 29 '16 at 14:25
$begingroup$
@individ : for $N=1,2,3,4$, the solutions $(m,n)$ are : $(33,217)$ , $(22641,961753)$ , $(263568049,55479822393)$ , $(45074835574129,41942086060150713)$.
$endgroup$
– uvdose
Feb 29 '16 at 14:25
5
5
$begingroup$
If (n+1) | m^2 + 1 and (m+ 1) | n^2 + 1 then (n + 1) | ( m + 1) ^2 - 2m and ((m+1) ^ 2) | n^4 + 2 (m^2) + 1 therefore ( n + 1) | (n^4 + 2( n^2) + 1)/ h - 2m , for some integer h . So ( n+ 1) | (n^4 + 2( n^2) + 1 - (2m h)) , so (n+1) | (4n^2 - (2 m h)) and ( n+1) | 2( m h +2) . 2| m iff 2| m , so if 2| n then ( n+1) | (m h +2) and similarly ( m+1) | ( n k+ 2) for some k an element of integers. I don't know if this helps....
$endgroup$
– 201044
Mar 15 '16 at 23:39
$begingroup$
If (n+1) | m^2 + 1 and (m+ 1) | n^2 + 1 then (n + 1) | ( m + 1) ^2 - 2m and ((m+1) ^ 2) | n^4 + 2 (m^2) + 1 therefore ( n + 1) | (n^4 + 2( n^2) + 1)/ h - 2m , for some integer h . So ( n+ 1) | (n^4 + 2( n^2) + 1 - (2m h)) , so (n+1) | (4n^2 - (2 m h)) and ( n+1) | 2( m h +2) . 2| m iff 2| m , so if 2| n then ( n+1) | (m h +2) and similarly ( m+1) | ( n k+ 2) for some k an element of integers. I don't know if this helps....
$endgroup$
– 201044
Mar 15 '16 at 23:39
|
show 5 more comments
4 Answers
4
active
oldest
votes
$begingroup$
Some further results along the lines of thought of @individ:
Suppose $p$ and $s$ are solutions to the Pell's equation:
$$-dcdot p^2+s^2=1$$
Then,
begin{align}
m &= acdot p^2+bcdot pq +ccdot q^2\
n &= acdot p^2-bcdot pq +ccdot q^2
end{align}
are solutions if $(a,b,c,d)$ are: (these are the only sets that I found using the computer)
begin{align}
(10,4,-2,-15)\
(39,12,-3,-65)\
end{align}
Sadly, the solutions are negative.
Here are some examples:
begin{align}
(m,n) &= (-6,-38) &(a,b,c,d,p,q)&=(10,4,-2,-15,1,4)\
(m,n) &= (-290,-2274) &(a,b,c,d,p,q)&=(10,4,-2,-15,8,31)\
(m,n) &= (-15171,-64707) &(a,b,c,d,p,q)&=(39,12,-3,-65,16,129)\
(m,n) &= (-1009692291,-4306907523) &(a,b,c,d,p,q)&=(39,12,-3,-65,4128,33281)\
(m,n) &= (-67207138138563,-286676378361411) &(a,b,c,d,p,q)&=(39,12,-3,-65,1065008,8586369)\
end{align}
P.S. I am also very curious how @individ thought of this parametrization.
answered Dec 31 '16 at 13:58
Yifan ZhuYifan Zhu
855
$endgroup$
add a comment |
$begingroup$
Trying again, now as answer instead of just a comment:
I now have a pdf containing the text that I could not place on the margin of Gauss' masterpiece. Anyone interested?
The pdf contains a full classification of (all) solutions and a procedure on how to generate them all.
simultaneouos_divisions_MSE.pdf
answered Dec 22 '18 at 15:47
Maestro13Maestro13
1,081724
$endgroup$
$begingroup$
why link pdf? why not just list answer here?
$endgroup$
– qwr
Jan 9 at 20:34
add a comment |
$begingroup$
You can record a similar system:
$$left{begin{aligned}&m^2+t^2=(n+t)z\&n^2+t^2=(m+t)kend{aligned}right.$$
Parametrization of solutions we write this.
$$m=q(3x-q)$$
$$n=2x^2-qx-q^2$$
$$t=3q^2-3xq+2x^2$$
$$z=5q^2-2qx+x^2$$
$$k=5q^2-8qx+4x^2$$
Consider a special case.
$$left{begin{aligned}&m^2+1=(n+1)z\&n^2+1=(m+1)kend{aligned}right.$$
Using the solutions of the equation Pell.
$$p^2-15s^2=1$$
Enough to know first, everything else will find a formula. $(p;s) - (4;1)$
$$p_2=4p+15s$$
$$s_2=p+4s$$
The solution then write.
$$m=-2p^2-4ps+10s^2$$
$$n=-2p^2+4ps+10s^2$$
$$z=8m+9-n$$
$$k=8n+9-m$$
These solutions are negative.
And a positive decision of the same are determined by the Pell equation.
$$p^2-65s^2=-1$$
Use the first solution. $(p;s) - (8;1)$
Next find the formula.
$$p_2=129p+1040s$$
$$s_2=16p+129s$$
Will make a replacement.
$$x=p^2+6ps+13s^2$$
$$y=p^2-6ps+13s^2$$
The decision record.
$$m=2x-1$$
$$n=2y-1$$
$$z=9x-2y+2$$
$$k=9y-2x+2$$
answered Dec 30 '16 at 8:44
individindivid
3,2521816
$endgroup$
$begingroup$
In your parametrization, with t = 1, we have $1 = 3q^2 - 3xq + 2x^2$, or equivalently $2x^2 - 3xq + (3q^2-1) = 0$. This equation can be solved in numbers only if the discriminant $9q^2 - 8(3q^2- 1) = 8 - 15q^2$ is a square, but this is obviously impossible. So, either your parametrization is OK and there is no solution, or your parametrization is wrong.
$endgroup$
– MikeTeX
Jan 14 at 15:50
$begingroup$
@MikeTeX there are several solutions. One of these decisions was mentioned. It's private. Other solutions need to be considered. Look. Several of them.
$endgroup$
– individ
Jan 14 at 16:18
add a comment |
$begingroup$
To solve this system of equations - it is necessary to solve the system.
$$left{begin{aligned}&m^2+t^2=(n+t)z\&n^2+t^2=(m+t)kend{aligned}right.$$
It is necessary to find a parameterization to figurirovallo Pell. It is possible for example to record this.
$$m=33643p^2pm5404ps+217s^2$$
$$n=5491p^2pm852ps+33s^2$$
$$t=s^2-153p^2$$
$$z=212041p^2pm34274ps+1385s^2$$
$$k=901p^2pm134ps+5s^2$$
We need a case of when. $t=s^2-153p^2=1$
Knowing the first decision $(s ; p ) - (2177;176)$
The rest can be found by the formula.
$$s=2177s+26928p$$
$$p=176s+2177p$$
Although this equation can be not enough. We need to find when there are multiple solutions.
answered Jan 18 '17 at 16:57
individindivid
3,2521816
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Some further results along the lines of thought of @individ:
Suppose $p$ and $s$ are solutions to the Pell's equation:
$$-dcdot p^2+s^2=1$$
Then,
begin{align}
m &= acdot p^2+bcdot pq +ccdot q^2\
n &= acdot p^2-bcdot pq +ccdot q^2
end{align}
are solutions if $(a,b,c,d)$ are: (these are the only sets that I found using the computer)
begin{align}
(10,4,-2,-15)\
(39,12,-3,-65)\
end{align}
Sadly, the solutions are negative.
Here are some examples:
begin{align}
(m,n) &= (-6,-38) &(a,b,c,d,p,q)&=(10,4,-2,-15,1,4)\
(m,n) &= (-290,-2274) &(a,b,c,d,p,q)&=(10,4,-2,-15,8,31)\
(m,n) &= (-15171,-64707) &(a,b,c,d,p,q)&=(39,12,-3,-65,16,129)\
(m,n) &= (-1009692291,-4306907523) &(a,b,c,d,p,q)&=(39,12,-3,-65,4128,33281)\
(m,n) &= (-67207138138563,-286676378361411) &(a,b,c,d,p,q)&=(39,12,-3,-65,1065008,8586369)\
end{align}
P.S. I am also very curious how @individ thought of this parametrization.
answered Dec 31 '16 at 13:58
Yifan ZhuYifan Zhu
855
$endgroup$
add a comment |
$begingroup$
Some further results along the lines of thought of @individ:
Suppose $p$ and $s$ are solutions to the Pell's equation:
$$-dcdot p^2+s^2=1$$
Then,
begin{align}
m &= acdot p^2+bcdot pq +ccdot q^2\
n &= acdot p^2-bcdot pq +ccdot q^2
end{align}
are solutions if $(a,b,c,d)$ are: (these are the only sets that I found using the computer)
begin{align}
(10,4,-2,-15)\
(39,12,-3,-65)\
end{align}
Sadly, the solutions are negative.
Here are some examples:
begin{align}
(m,n) &= (-6,-38) &(a,b,c,d,p,q)&=(10,4,-2,-15,1,4)\
(m,n) &= (-290,-2274) &(a,b,c,d,p,q)&=(10,4,-2,-15,8,31)\
(m,n) &= (-15171,-64707) &(a,b,c,d,p,q)&=(39,12,-3,-65,16,129)\
(m,n) &= (-1009692291,-4306907523) &(a,b,c,d,p,q)&=(39,12,-3,-65,4128,33281)\
(m,n) &= (-67207138138563,-286676378361411) &(a,b,c,d,p,q)&=(39,12,-3,-65,1065008,8586369)\
end{align}
P.S. I am also very curious how @individ thought of this parametrization.
answered Dec 31 '16 at 13:58
Yifan ZhuYifan Zhu
855
$endgroup$
add a comment |
$begingroup$
Some further results along the lines of thought of @individ:
Suppose $p$ and $s$ are solutions to the Pell's equation:
$$-dcdot p^2+s^2=1$$
Then,
begin{align}
m &= acdot p^2+bcdot pq +ccdot q^2\
n &= acdot p^2-bcdot pq +ccdot q^2
end{align}
are solutions if $(a,b,c,d)$ are: (these are the only sets that I found using the computer)
begin{align}
(10,4,-2,-15)\
(39,12,-3,-65)\
end{align}
Sadly, the solutions are negative.
Here are some examples:
begin{align}
(m,n) &= (-6,-38) &(a,b,c,d,p,q)&=(10,4,-2,-15,1,4)\
(m,n) &= (-290,-2274) &(a,b,c,d,p,q)&=(10,4,-2,-15,8,31)\
(m,n) &= (-15171,-64707) &(a,b,c,d,p,q)&=(39,12,-3,-65,16,129)\
(m,n) &= (-1009692291,-4306907523) &(a,b,c,d,p,q)&=(39,12,-3,-65,4128,33281)\
(m,n) &= (-67207138138563,-286676378361411) &(a,b,c,d,p,q)&=(39,12,-3,-65,1065008,8586369)\
end{align}
P.S. I am also very curious how @individ thought of this parametrization.
answered Dec 31 '16 at 13:58
Yifan ZhuYifan Zhu
855
$endgroup$
Some further results along the lines of thought of @individ:
Suppose $p$ and $s$ are solutions to the Pell's equation:
$$-dcdot p^2+s^2=1$$
Then,
begin{align}
m &= acdot p^2+bcdot pq +ccdot q^2\
n &= acdot p^2-bcdot pq +ccdot q^2
end{align}
are solutions if $(a,b,c,d)$ are: (these are the only sets that I found using the computer)
begin{align}
(10,4,-2,-15)\
(39,12,-3,-65)\
end{align}
Sadly, the solutions are negative.
Here are some examples:
begin{align}
(m,n) &= (-6,-38) &(a,b,c,d,p,q)&=(10,4,-2,-15,1,4)\
(m,n) &= (-290,-2274) &(a,b,c,d,p,q)&=(10,4,-2,-15,8,31)\
(m,n) &= (-15171,-64707) &(a,b,c,d,p,q)&=(39,12,-3,-65,16,129)\
(m,n) &= (-1009692291,-4306907523) &(a,b,c,d,p,q)&=(39,12,-3,-65,4128,33281)\
(m,n) &= (-67207138138563,-286676378361411) &(a,b,c,d,p,q)&=(39,12,-3,-65,1065008,8586369)\
end{align}
P.S. I am also very curious how @individ thought of this parametrization.
answered Dec 31 '16 at 13:58
Yifan ZhuYifan Zhu
855
answered Dec 31 '16 at 13:58
Yifan ZhuYifan Zhu
855
answered Dec 31 '16 at 13:58
Yifan ZhuYifan Zhu
855
answered Dec 31 '16 at 13:58
Yifan ZhuYifan Zhu
855
855
add a comment |
add a comment |
$begingroup$
Trying again, now as answer instead of just a comment:
I now have a pdf containing the text that I could not place on the margin of Gauss' masterpiece. Anyone interested?
The pdf contains a full classification of (all) solutions and a procedure on how to generate them all.
simultaneouos_divisions_MSE.pdf
answered Dec 22 '18 at 15:47
Maestro13Maestro13
1,081724
$endgroup$
$begingroup$
why link pdf? why not just list answer here?
$endgroup$
– qwr
Jan 9 at 20:34
add a comment |
$begingroup$
Trying again, now as answer instead of just a comment:
I now have a pdf containing the text that I could not place on the margin of Gauss' masterpiece. Anyone interested?
The pdf contains a full classification of (all) solutions and a procedure on how to generate them all.
simultaneouos_divisions_MSE.pdf
answered Dec 22 '18 at 15:47
Maestro13Maestro13
1,081724
$endgroup$
$begingroup$
why link pdf? why not just list answer here?
$endgroup$
– qwr
Jan 9 at 20:34
add a comment |
$begingroup$
Trying again, now as answer instead of just a comment:
I now have a pdf containing the text that I could not place on the margin of Gauss' masterpiece. Anyone interested?
The pdf contains a full classification of (all) solutions and a procedure on how to generate them all.
simultaneouos_divisions_MSE.pdf
answered Dec 22 '18 at 15:47
Maestro13Maestro13
1,081724
$endgroup$
Trying again, now as answer instead of just a comment:
I now have a pdf containing the text that I could not place on the margin of Gauss' masterpiece. Anyone interested?
The pdf contains a full classification of (all) solutions and a procedure on how to generate them all.
simultaneouos_divisions_MSE.pdf
answered Dec 22 '18 at 15:47
Maestro13Maestro13
1,081724
edited Dec 22 '18 at 21:58
answered Dec 22 '18 at 15:47
Maestro13Maestro13
1,081724
answered Dec 22 '18 at 15:47
Maestro13Maestro13
1,081724
answered Dec 22 '18 at 15:47
Maestro13Maestro13
1,081724
1,081724
$begingroup$
why link pdf? why not just list answer here?
$endgroup$
– qwr
Jan 9 at 20:34
add a comment |
$begingroup$
why link pdf? why not just list answer here?
$endgroup$
– qwr
Jan 9 at 20:34
$begingroup$
why link pdf? why not just list answer here?
$endgroup$
– qwr
Jan 9 at 20:34
$begingroup$
why link pdf? why not just list answer here?
$endgroup$
– qwr
Jan 9 at 20:34
add a comment |
$begingroup$
You can record a similar system:
$$left{begin{aligned}&m^2+t^2=(n+t)z\&n^2+t^2=(m+t)kend{aligned}right.$$
Parametrization of solutions we write this.
$$m=q(3x-q)$$
$$n=2x^2-qx-q^2$$
$$t=3q^2-3xq+2x^2$$
$$z=5q^2-2qx+x^2$$
$$k=5q^2-8qx+4x^2$$
Consider a special case.
$$left{begin{aligned}&m^2+1=(n+1)z\&n^2+1=(m+1)kend{aligned}right.$$
Using the solutions of the equation Pell.
$$p^2-15s^2=1$$
Enough to know first, everything else will find a formula. $(p;s) - (4;1)$
$$p_2=4p+15s$$
$$s_2=p+4s$$
The solution then write.
$$m=-2p^2-4ps+10s^2$$
$$n=-2p^2+4ps+10s^2$$
$$z=8m+9-n$$
$$k=8n+9-m$$
These solutions are negative.
And a positive decision of the same are determined by the Pell equation.
$$p^2-65s^2=-1$$
Use the first solution. $(p;s) - (8;1)$
Next find the formula.
$$p_2=129p+1040s$$
$$s_2=16p+129s$$
Will make a replacement.
$$x=p^2+6ps+13s^2$$
$$y=p^2-6ps+13s^2$$
The decision record.
$$m=2x-1$$
$$n=2y-1$$
$$z=9x-2y+2$$
$$k=9y-2x+2$$
answered Dec 30 '16 at 8:44
individindivid
3,2521816
$endgroup$
$begingroup$
In your parametrization, with t = 1, we have $1 = 3q^2 - 3xq + 2x^2$, or equivalently $2x^2 - 3xq + (3q^2-1) = 0$. This equation can be solved in numbers only if the discriminant $9q^2 - 8(3q^2- 1) = 8 - 15q^2$ is a square, but this is obviously impossible. So, either your parametrization is OK and there is no solution, or your parametrization is wrong.
$endgroup$
– MikeTeX
Jan 14 at 15:50
$begingroup$
@MikeTeX there are several solutions. One of these decisions was mentioned. It's private. Other solutions need to be considered. Look. Several of them.
$endgroup$
– individ
Jan 14 at 16:18
add a comment |
$begingroup$
You can record a similar system:
$$left{begin{aligned}&m^2+t^2=(n+t)z\&n^2+t^2=(m+t)kend{aligned}right.$$
Parametrization of solutions we write this.
$$m=q(3x-q)$$
$$n=2x^2-qx-q^2$$
$$t=3q^2-3xq+2x^2$$
$$z=5q^2-2qx+x^2$$
$$k=5q^2-8qx+4x^2$$
Consider a special case.
$$left{begin{aligned}&m^2+1=(n+1)z\&n^2+1=(m+1)kend{aligned}right.$$
Using the solutions of the equation Pell.
$$p^2-15s^2=1$$
Enough to know first, everything else will find a formula. $(p;s) - (4;1)$
$$p_2=4p+15s$$
$$s_2=p+4s$$
The solution then write.
$$m=-2p^2-4ps+10s^2$$
$$n=-2p^2+4ps+10s^2$$
$$z=8m+9-n$$
$$k=8n+9-m$$
These solutions are negative.
And a positive decision of the same are determined by the Pell equation.
$$p^2-65s^2=-1$$
Use the first solution. $(p;s) - (8;1)$
Next find the formula.
$$p_2=129p+1040s$$
$$s_2=16p+129s$$
Will make a replacement.
$$x=p^2+6ps+13s^2$$
$$y=p^2-6ps+13s^2$$
The decision record.
$$m=2x-1$$
$$n=2y-1$$
$$z=9x-2y+2$$
$$k=9y-2x+2$$
answered Dec 30 '16 at 8:44
individindivid
3,2521816
$endgroup$
$begingroup$
In your parametrization, with t = 1, we have $1 = 3q^2 - 3xq + 2x^2$, or equivalently $2x^2 - 3xq + (3q^2-1) = 0$. This equation can be solved in numbers only if the discriminant $9q^2 - 8(3q^2- 1) = 8 - 15q^2$ is a square, but this is obviously impossible. So, either your parametrization is OK and there is no solution, or your parametrization is wrong.
$endgroup$
– MikeTeX
Jan 14 at 15:50
$begingroup$
@MikeTeX there are several solutions. One of these decisions was mentioned. It's private. Other solutions need to be considered. Look. Several of them.
$endgroup$
– individ
Jan 14 at 16:18
add a comment |
$begingroup$
You can record a similar system:
$$left{begin{aligned}&m^2+t^2=(n+t)z\&n^2+t^2=(m+t)kend{aligned}right.$$
Parametrization of solutions we write this.
$$m=q(3x-q)$$
$$n=2x^2-qx-q^2$$
$$t=3q^2-3xq+2x^2$$
$$z=5q^2-2qx+x^2$$
$$k=5q^2-8qx+4x^2$$
Consider a special case.
$$left{begin{aligned}&m^2+1=(n+1)z\&n^2+1=(m+1)kend{aligned}right.$$
Using the solutions of the equation Pell.
$$p^2-15s^2=1$$
Enough to know first, everything else will find a formula. $(p;s) - (4;1)$
$$p_2=4p+15s$$
$$s_2=p+4s$$
The solution then write.
$$m=-2p^2-4ps+10s^2$$
$$n=-2p^2+4ps+10s^2$$
$$z=8m+9-n$$
$$k=8n+9-m$$
These solutions are negative.
And a positive decision of the same are determined by the Pell equation.
$$p^2-65s^2=-1$$
Use the first solution. $(p;s) - (8;1)$
Next find the formula.
$$p_2=129p+1040s$$
$$s_2=16p+129s$$
Will make a replacement.
$$x=p^2+6ps+13s^2$$
$$y=p^2-6ps+13s^2$$
The decision record.
$$m=2x-1$$
$$n=2y-1$$
$$z=9x-2y+2$$
$$k=9y-2x+2$$
answered Dec 30 '16 at 8:44
individindivid
3,2521816
$endgroup$
You can record a similar system:
$$left{begin{aligned}&m^2+t^2=(n+t)z\&n^2+t^2=(m+t)kend{aligned}right.$$
Parametrization of solutions we write this.
$$m=q(3x-q)$$
$$n=2x^2-qx-q^2$$
$$t=3q^2-3xq+2x^2$$
$$z=5q^2-2qx+x^2$$
$$k=5q^2-8qx+4x^2$$
Consider a special case.
$$left{begin{aligned}&m^2+1=(n+1)z\&n^2+1=(m+1)kend{aligned}right.$$
Using the solutions of the equation Pell.
$$p^2-15s^2=1$$
Enough to know first, everything else will find a formula. $(p;s) - (4;1)$
$$p_2=4p+15s$$
$$s_2=p+4s$$
The solution then write.
$$m=-2p^2-4ps+10s^2$$
$$n=-2p^2+4ps+10s^2$$
$$z=8m+9-n$$
$$k=8n+9-m$$
These solutions are negative.
And a positive decision of the same are determined by the Pell equation.
$$p^2-65s^2=-1$$
Use the first solution. $(p;s) - (8;1)$
Next find the formula.
$$p_2=129p+1040s$$
$$s_2=16p+129s$$
Will make a replacement.
$$x=p^2+6ps+13s^2$$
$$y=p^2-6ps+13s^2$$
The decision record.
$$m=2x-1$$
$$n=2y-1$$
$$z=9x-2y+2$$
$$k=9y-2x+2$$
answered Dec 30 '16 at 8:44
individindivid
3,2521816
edited Jan 14 '17 at 12:45
answered Dec 30 '16 at 8:44
individindivid
3,2521816
answered Dec 30 '16 at 8:44
individindivid
3,2521816
answered Dec 30 '16 at 8:44
individindivid
3,2521816
3,2521816
$begingroup$
In your parametrization, with t = 1, we have $1 = 3q^2 - 3xq + 2x^2$, or equivalently $2x^2 - 3xq + (3q^2-1) = 0$. This equation can be solved in numbers only if the discriminant $9q^2 - 8(3q^2- 1) = 8 - 15q^2$ is a square, but this is obviously impossible. So, either your parametrization is OK and there is no solution, or your parametrization is wrong.
$endgroup$
– MikeTeX
Jan 14 at 15:50
$begingroup$
@MikeTeX there are several solutions. One of these decisions was mentioned. It's private. Other solutions need to be considered. Look. Several of them.
$endgroup$
– individ
Jan 14 at 16:18
add a comment |
$begingroup$
In your parametrization, with t = 1, we have $1 = 3q^2 - 3xq + 2x^2$, or equivalently $2x^2 - 3xq + (3q^2-1) = 0$. This equation can be solved in numbers only if the discriminant $9q^2 - 8(3q^2- 1) = 8 - 15q^2$ is a square, but this is obviously impossible. So, either your parametrization is OK and there is no solution, or your parametrization is wrong.
$endgroup$
– MikeTeX
Jan 14 at 15:50
$begingroup$
@MikeTeX there are several solutions. One of these decisions was mentioned. It's private. Other solutions need to be considered. Look. Several of them.
$endgroup$
– individ
Jan 14 at 16:18
$begingroup$
In your parametrization, with t = 1, we have $1 = 3q^2 - 3xq + 2x^2$, or equivalently $2x^2 - 3xq + (3q^2-1) = 0$. This equation can be solved in numbers only if the discriminant $9q^2 - 8(3q^2- 1) = 8 - 15q^2$ is a square, but this is obviously impossible. So, either your parametrization is OK and there is no solution, or your parametrization is wrong.
$endgroup$
– MikeTeX
Jan 14 at 15:50
$begingroup$
In your parametrization, with t = 1, we have $1 = 3q^2 - 3xq + 2x^2$, or equivalently $2x^2 - 3xq + (3q^2-1) = 0$. This equation can be solved in numbers only if the discriminant $9q^2 - 8(3q^2- 1) = 8 - 15q^2$ is a square, but this is obviously impossible. So, either your parametrization is OK and there is no solution, or your parametrization is wrong.
$endgroup$
– MikeTeX
Jan 14 at 15:50
$begingroup$
@MikeTeX there are several solutions. One of these decisions was mentioned. It's private. Other solutions need to be considered. Look. Several of them.
$endgroup$
– individ
Jan 14 at 16:18
$begingroup$
@MikeTeX there are several solutions. One of these decisions was mentioned. It's private. Other solutions need to be considered. Look. Several of them.
$endgroup$
– individ
Jan 14 at 16:18
add a comment |
$begingroup$
To solve this system of equations - it is necessary to solve the system.
$$left{begin{aligned}&m^2+t^2=(n+t)z\&n^2+t^2=(m+t)kend{aligned}right.$$
It is necessary to find a parameterization to figurirovallo Pell. It is possible for example to record this.
$$m=33643p^2pm5404ps+217s^2$$
$$n=5491p^2pm852ps+33s^2$$
$$t=s^2-153p^2$$
$$z=212041p^2pm34274ps+1385s^2$$
$$k=901p^2pm134ps+5s^2$$
We need a case of when. $t=s^2-153p^2=1$
Knowing the first decision $(s ; p ) - (2177;176)$
The rest can be found by the formula.
$$s=2177s+26928p$$
$$p=176s+2177p$$
Although this equation can be not enough. We need to find when there are multiple solutions.
answered Jan 18 '17 at 16:57
individindivid
3,2521816
$endgroup$
add a comment |
$begingroup$
To solve this system of equations - it is necessary to solve the system.
$$left{begin{aligned}&m^2+t^2=(n+t)z\&n^2+t^2=(m+t)kend{aligned}right.$$
It is necessary to find a parameterization to figurirovallo Pell. It is possible for example to record this.
$$m=33643p^2pm5404ps+217s^2$$
$$n=5491p^2pm852ps+33s^2$$
$$t=s^2-153p^2$$
$$z=212041p^2pm34274ps+1385s^2$$
$$k=901p^2pm134ps+5s^2$$
We need a case of when. $t=s^2-153p^2=1$
Knowing the first decision $(s ; p ) - (2177;176)$
The rest can be found by the formula.
$$s=2177s+26928p$$
$$p=176s+2177p$$
Although this equation can be not enough. We need to find when there are multiple solutions.
answered Jan 18 '17 at 16:57
individindivid
3,2521816
$endgroup$
add a comment |
$begingroup$
To solve this system of equations - it is necessary to solve the system.
$$left{begin{aligned}&m^2+t^2=(n+t)z\&n^2+t^2=(m+t)kend{aligned}right.$$
It is necessary to find a parameterization to figurirovallo Pell. It is possible for example to record this.
$$m=33643p^2pm5404ps+217s^2$$
$$n=5491p^2pm852ps+33s^2$$
$$t=s^2-153p^2$$
$$z=212041p^2pm34274ps+1385s^2$$
$$k=901p^2pm134ps+5s^2$$
We need a case of when. $t=s^2-153p^2=1$
Knowing the first decision $(s ; p ) - (2177;176)$
The rest can be found by the formula.
$$s=2177s+26928p$$
$$p=176s+2177p$$
Although this equation can be not enough. We need to find when there are multiple solutions.
answered Jan 18 '17 at 16:57
individindivid
3,2521816
$endgroup$
To solve this system of equations - it is necessary to solve the system.
$$left{begin{aligned}&m^2+t^2=(n+t)z\&n^2+t^2=(m+t)kend{aligned}right.$$
It is necessary to find a parameterization to figurirovallo Pell. It is possible for example to record this.
$$m=33643p^2pm5404ps+217s^2$$
$$n=5491p^2pm852ps+33s^2$$
$$t=s^2-153p^2$$
$$z=212041p^2pm34274ps+1385s^2$$
$$k=901p^2pm134ps+5s^2$$
We need a case of when. $t=s^2-153p^2=1$
Knowing the first decision $(s ; p ) - (2177;176)$
The rest can be found by the formula.
$$s=2177s+26928p$$
$$p=176s+2177p$$
Although this equation can be not enough. We need to find when there are multiple solutions.
answered Jan 18 '17 at 16:57
individindivid
3,2521816
answered Jan 18 '17 at 16:57
individindivid
3,2521816
answered Jan 18 '17 at 16:57
individindivid
3,2521816
answered Jan 18 '17 at 16:57
individindivid
3,2521816
3,2521816
add a comment |
add a comment |
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12
$begingroup$
Could you write down your proof for the infinite number of solutions? It may help everyone.
$endgroup$
– user37238
Feb 25 '16 at 8:22
3
$begingroup$
If you understand French, there is a long discussion (partly numerical, partly theoretical) here
$endgroup$
– Giovanni Resta
Feb 25 '16 at 8:28
2
$begingroup$
Yes, I understand French: in fact, I participated in this discussion. One of the participants (serge17) managed to find a method to build "big" solutions $(m,n)$ : Using his method, we can prove that there are infinitely many solutions.
$endgroup$
– uvdose
Feb 25 '16 at 8:37
4
$begingroup$
@individ : for $N=1,2,3,4$, the solutions $(m,n)$ are : $(33,217)$ , $(22641,961753)$ , $(263568049,55479822393)$ , $(45074835574129,41942086060150713)$.
$endgroup$
– uvdose
Feb 29 '16 at 14:25
5
$begingroup$
If (n+1) | m^2 + 1 and (m+ 1) | n^2 + 1 then (n + 1) | ( m + 1) ^2 - 2m and ((m+1) ^ 2) | n^4 + 2 (m^2) + 1 therefore ( n + 1) | (n^4 + 2( n^2) + 1)/ h - 2m , for some integer h . So ( n+ 1) | (n^4 + 2( n^2) + 1 - (2m h)) , so (n+1) | (4n^2 - (2 m h)) and ( n+1) | 2( m h +2) . 2| m iff 2| m , so if 2| n then ( n+1) | (m h +2) and similarly ( m+1) | ( n k+ 2) for some k an element of integers. I don't know if this helps....
$endgroup$
– 201044
Mar 15 '16 at 23:39