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From: The Pumpster on 9 Aug 2010 09:04
On Aug 9, 5:32 am, "alainvergh...(a)gmail.com" <alainvergh...(a)gmail.com>
wrote:
> On 9 août, 00:47, The Pumpster <pumpledumplek...(a)gmail.com> wrote:
>
>
>
>
>
> > On Aug 8, 3:43 pm, quasi <qu...(a)null.set> wrote:
>
> > > On Sun, 8 Aug 2010 09:55:52 -0700 (PDT), The Pumpster
>
> > > <pumpledumplek...(a)gmail.com> wrote:
> > > >On Aug 8, 5:54 am, "alainvergh...(a)gmail.com" <alainvergh...(a)gmail.com>
> > > >wrote:
> > > >> Good morning,
>
> > > >> I need your help to solve this problem:
>
> > > >> v and z positive integer , v>z
> > > >> how can we get v^2+3zv and v^2-zv also squares?
>
> > > >> Thanks for any ideas,
>
> > > >> Alain
>
> > > >The corresponding elliptic curve (of conductor 24)
> > > >has rank 0. Checking the torsion points leads only to
> > > >the obvious points with |v| and |z| <=1, so there
> > > >are no solutions to your problem.
>
> > > In fact, there are infinitely many solutions.
>
> > > One example is v=49, z=40.
>
> > > Of course, any positive integer multiple of a solution pair is also a
> > > solution pair, so it suffices to consider primitive solutions, i.e.,
> > > solution pairs (v,z) with gcd(v,z)=1. But there are also infinitely
> > > many of those.
>
> > > Assume (v,z) is a primitive solution. Then gcd(v,z)=1 implies
> > > gcd(v,v-z)=1, so v (v-z) a square implies that both v and v-z are
> > > squares. Then, since v is a square and v (v+3z) is a square , it
> > > follows that v+3z is also a square. Thus, write
>
> > >     v + 3z = a^2
> > >     v - z = b^2
>
> > > where a,b are either both odd or both even.
>
> > > Then v = (a^2 + 3b^2)/4 and z = (a^2 - b^2)/4
>
> > > If a,b are both even, then v,z would both be even, hence a,b must both
> > > be odd.
>
> > > Since v is a square, write v = c^2, so we get
>
> > >     a^2 + 3b^2 = 4c^2
>
> > > Given any positive integer solution a,b,c to the above where a > b,
> > > gcd(a,b) = 1, and a,b both odd, we get a primitive solution v,z
> > > satisfying the original requirements, and conversely, this gives all
> > > primitive solutions.
>
> > > To solving the above equation start with
>
> > >    3b^2 = (2c+a) (2c-a).
>
> > > Claim 2c+a and 2c-a are relatively prime. Clearly they are both odd.
> > > Suppose that they have a common prime factor p. Then p is odd.
>
> > > Then p | ((2c+a) - (2c-a)) => p | 2a => p | a.
>
> > > Also, p^2 |(2c+a) (2c-a) => p^2 | 3b^2 => p | b,
>
> > > contrary to the assumption that gcd(a,b) = 1.
>
> > > Thus, 2c+a and 2c-a are relatively prime, as claimed. Then
>
> > >    3b^2 = (2c+a) (2c-a)
>
> > > implies either
>
> > >    2c + a = 3s^2
> > >    2c - a = t^2
>
> > > or
>
> > >    2c + a = s^2
> > >    2c - a = 3t^2
>
> > > for some positive integers s,t, both odd, with gcd(s,t)=1.
>
> > > Solving for a,c, then substituting to get b, and finally, substituting
> > > to get v,z yields 2 classes of solutions.
>
> > > Class #1:
>
> > >    v = ((3s^2 + t^2)/4)^2
> > >    z = (9s^4 - 10s^2 t^2 + t^4)/16
>
> > > where s,t are both odd positive integers, with gcd(s,t) = 1 and either
> > > t<s or t>3s.
>
> > > Class #2:
>
> > >    v = ((3s^2 + t^2)/4)^2
> > >    z = (s^4 - 10s^2 t^2 + 9 t^4)/16
>
> > > where s,t are both odd positive integers, with gcd(s,t) = 1 and either
> > > s>t or s<3t.
>
> > > I haven't carefully checked the algebra -- hopefully I haven't made
> > > errors. However I'm fairly sure the basic idea is correct, and I'm
> > > absolutely sure that there really are infinitely many primitive
> > > solutions.
>
> > > quasi
>
> > ...ah but if you mistyped the original equations into Magma in exactly
> > the same way that I did, you could have been as incorrect as I was! I
> > think
> > I did the "same" problem with v^2+3z^2 and v^2-z^2.....
>
> > You are of course correct. The question is one of pythagorean triples
> > (roughly) as you say...
>
> > (hanging head in shame) de P- Masquer le texte des messages précédents -
>
> > - Afficher le texte des messages précédents -
>
> Bonjour,
>
> Thanks for replying.
>
> I feel sorry , I've made a mistake :reading back
> my problem I've seen I expected three squares:
> v^2 , v^2-zv+2z^2 and v^2+3zv
> Okay your formula works for v^2+3zv a square,
>
> Best regards,
>
> Alain

Well, this problem really does lead to something of
genus 1 instead of 0! It corresponds to an elliptic
curve of rank 1 over Q, and conductor 1176 (again, this
is all assuming that I can type), so it seems that
there are infinitely many solutions of the desired shape
(such as v=529, z=440), which can be expressed recursively...

de P
From: alainverghote on 9 Aug 2010 11:32
On 9 août, 15:04, The Pumpster <pumpledumplek...(a)gmail.com> wrote:
> On Aug 9, 5:32 am, "alainvergh...(a)gmail.com" <alainvergh...(a)gmail.com>
> wrote:
>
>
>
>
>
> > On 9 août, 00:47, The Pumpster <pumpledumplek...(a)gmail.com> wrote:
>
> > > On Aug 8, 3:43 pm, quasi <qu...(a)null.set> wrote:
>
> > > > On Sun, 8 Aug 2010 09:55:52 -0700 (PDT), The Pumpster
>
> > > > <pumpledumplek...(a)gmail.com> wrote:
> > > > >On Aug 8, 5:54 am, "alainvergh...(a)gmail.com" <alainvergh...(a)gmail.com>
> > > > >wrote:
> > > > >> Good morning,
>
> > > > >> I need your help to solve this problem:
>
> > > > >> v and z positive integer , v>z
> > > > >> how can we get v^2+3zv and v^2-zv also squares?
>
> > > > >> Thanks for any ideas,
>
> > > > >> Alain
>
> > > > >The corresponding elliptic curve (of conductor 24)
> > > > >has rank 0. Checking the torsion points leads only to
> > > > >the obvious points with |v| and |z| <=1, so there
> > > > >are no solutions to your problem.
>
> > > > In fact, there are infinitely many solutions.
>
> > > > One example is v=49, z=40.
>
> > > > Of course, any positive integer multiple of a solution pair is also a
> > > > solution pair, so it suffices to consider primitive solutions, i.e.,
> > > > solution pairs (v,z) with gcd(v,z)=1. But there are also infinitely
> > > > many of those.
>
> > > > Assume (v,z) is a primitive solution. Then gcd(v,z)=1 implies
> > > > gcd(v,v-z)=1, so v (v-z) a square implies that both v and v-z are
> > > > squares. Then, since v is a square and v (v+3z) is a square , it
> > > > follows that v+3z is also a square. Thus, write
>
> > > >     v + 3z = a^2
> > > >     v - z = b^2
>
> > > > where a,b are either both odd or both even.
>
> > > > Then v = (a^2 + 3b^2)/4 and z = (a^2 - b^2)/4
>
> > > > If a,b are both even, then v,z would both be even, hence a,b must both
> > > > be odd.
>
> > > > Since v is a square, write v = c^2, so we get
>
> > > >     a^2 + 3b^2 = 4c^2
>
> > > > Given any positive integer solution a,b,c to the above where a > b,
> > > > gcd(a,b) = 1, and a,b both odd, we get a primitive solution v,z
> > > > satisfying the original requirements, and conversely, this gives all
> > > > primitive solutions.
>
> > > > To solving the above equation start with
>
> > > >    3b^2 = (2c+a) (2c-a).
>
> > > > Claim 2c+a and 2c-a are relatively prime. Clearly they are both odd..
> > > > Suppose that they have a common prime factor p. Then p is odd.
>
> > > > Then p | ((2c+a) - (2c-a)) => p | 2a => p | a.
>
> > > > Also, p^2 |(2c+a) (2c-a) => p^2 | 3b^2 => p | b,
>
> > > > contrary to the assumption that gcd(a,b) = 1.
>
> > > > Thus, 2c+a and 2c-a are relatively prime, as claimed. Then
>
> > > >    3b^2 = (2c+a) (2c-a)
>
> > > > implies either
>
> > > >    2c + a = 3s^2
> > > >    2c - a = t^2
>
> > > > or
>
> > > >    2c + a = s^2
> > > >    2c - a = 3t^2
>
> > > > for some positive integers s,t, both odd, with gcd(s,t)=1.
>
> > > > Solving for a,c, then substituting to get b, and finally, substituting
> > > > to get v,z yields 2 classes of solutions.
>
> > > > Class #1:
>
> > > >    v = ((3s^2 + t^2)/4)^2
> > > >    z = (9s^4 - 10s^2 t^2 + t^4)/16
>
> > > > where s,t are both odd positive integers, with gcd(s,t) = 1 and either
> > > > t<s or t>3s.
>
> > > > Class #2:
>
> > > >    v = ((3s^2 + t^2)/4)^2
> > > >    z = (s^4 - 10s^2 t^2 + 9 t^4)/16
>
> > > > where s,t are both odd positive integers, with gcd(s,t) = 1 and either
> > > > s>t or s<3t.
>
> > > > I haven't carefully checked the algebra -- hopefully I haven't made
> > > > errors. However I'm fairly sure the basic idea is correct, and I'm
> > > > absolutely sure that there really are infinitely many primitive
> > > > solutions.
>
> > > > quasi
>
> > > ...ah but if you mistyped the original equations into Magma in exactly
> > > the same way that I did, you could have been as incorrect as I was! I
> > > think
> > > I did the "same" problem with v^2+3z^2 and v^2-z^2.....
>
> > > You are of course correct. The question is one of pythagorean triples
> > > (roughly) as you say...
>
> > > (hanging head in shame) de P- Masquer le texte des messages précédents -
>
> > > - Afficher le texte des messages précédents -
>
> > Bonjour,
>
> > Thanks for replying.
>
> > I feel sorry , I've made a mistake :reading back
> > my problem I've seen I expected three squares:
> > v^2 , v^2-zv+2z^2 and v^2+3zv
> > Okay your formula works for v^2+3zv a square,
>
> > Best regards,
>
> > Alain
>
> Well, this problem really does lead to something of
> genus 1 instead of 0! It corresponds to an elliptic
> curve of rank 1 over Q, and conductor 1176 (again, this
> is all assuming that I can type), so it seems that
> there are infinitely many solutions of the desired shape
> (such as v=529, z=440), which can be expressed recursively...
>
> de P- Masquer le texte des messages précédents -
>
> - Afficher le texte des messages précédents -

Thanks a lot for your reply,

Would you mind giving me some recursive expression,
f.i from the given values v=529, z=440 ...

Alain
From: The Pumpster on 9 Aug 2010 23:24
On Aug 9, 10:32 am, "alainvergh...(a)gmail.com"
<alainvergh...(a)gmail.com> wrote:
> On 9 août, 15:04, The Pumpster <pumpledumplek...(a)gmail.com> wrote:
>
>
>
>
>
> > On Aug 9, 5:32 am, "alainvergh...(a)gmail.com" <alainvergh...(a)gmail.com>
> > wrote:
>
> > > On 9 août, 00:47, The Pumpster <pumpledumplek...(a)gmail.com> wrote:
>
> > > > On Aug 8, 3:43 pm, quasi <qu...(a)null.set> wrote:
>
> > > > > On Sun, 8 Aug 2010 09:55:52 -0700 (PDT), The Pumpster
>
> > > > > <pumpledumplek...(a)gmail.com> wrote:
> > > > > >On Aug 8, 5:54 am, "alainvergh...(a)gmail.com" <alainvergh...(a)gmail.com>
> > > > > >wrote:
> > > > > >> Good morning,
>
> > > > > >> I need your help to solve this problem:
>
> > > > > >> v and z positive integer , v>z
> > > > > >> how can we get v^2+3zv and v^2-zv also squares?
>
> > > > > >> Thanks for any ideas,
>
> > > > > >> Alain
>
> > > > > >The corresponding elliptic curve (of conductor 24)
> > > > > >has rank 0. Checking the torsion points leads only to
> > > > > >the obvious points with |v| and |z| <=1, so there
> > > > > >are no solutions to your problem.
>
> > > > > In fact, there are infinitely many solutions.
>
> > > > > One example is v=49, z=40.
>
> > > > > Of course, any positive integer multiple of a solution pair is also a
> > > > > solution pair, so it suffices to consider primitive solutions, i.e.,
> > > > > solution pairs (v,z) with gcd(v,z)=1. But there are also infinitely
> > > > > many of those.
>
> > > > > Assume (v,z) is a primitive solution. Then gcd(v,z)=1 implies
> > > > > gcd(v,v-z)=1, so v (v-z) a square implies that both v and v-z are
> > > > > squares. Then, since v is a square and v (v+3z) is a square , it
> > > > > follows that v+3z is also a square. Thus, write
>
> > > > >     v + 3z = a^2
> > > > >     v - z = b^2
>
> > > > > where a,b are either both odd or both even.
>
> > > > > Then v = (a^2 + 3b^2)/4 and z = (a^2 - b^2)/4
>
> > > > > If a,b are both even, then v,z would both be even, hence a,b must both
> > > > > be odd.
>
> > > > > Since v is a square, write v = c^2, so we get
>
> > > > >     a^2 + 3b^2 = 4c^2
>
> > > > > Given any positive integer solution a,b,c to the above where a > b,
> > > > > gcd(a,b) = 1, and a,b both odd, we get a primitive solution v,z
> > > > > satisfying the original requirements, and conversely, this gives all
> > > > > primitive solutions.
>
> > > > > To solving the above equation start with
>
> > > > >    3b^2 = (2c+a) (2c-a).
>
> > > > > Claim 2c+a and 2c-a are relatively prime. Clearly they are both odd.
> > > > > Suppose that they have a common prime factor p. Then p is odd.
>
> > > > > Then p | ((2c+a) - (2c-a)) => p | 2a => p | a.
>
> > > > > Also, p^2 |(2c+a) (2c-a) => p^2 | 3b^2 => p | b,
>
> > > > > contrary to the assumption that gcd(a,b) = 1.
>
> > > > > Thus, 2c+a and 2c-a are relatively prime, as claimed. Then
>
> > > > >    3b^2 = (2c+a) (2c-a)
>
> > > > > implies either
>
> > > > >    2c + a = 3s^2
> > > > >    2c - a = t^2
>
> > > > > or
>
> > > > >    2c + a = s^2
> > > > >    2c - a = 3t^2
>
> > > > > for some positive integers s,t, both odd, with gcd(s,t)=1.
>
> > > > > Solving for a,c, then substituting to get b, and finally, substituting
> > > > > to get v,z yields 2 classes of solutions.
>
> > > > > Class #1:
>
> > > > >    v = ((3s^2 + t^2)/4)^2
> > > > >    z = (9s^4 - 10s^2 t^2 + t^4)/16
>
> > > > > where s,t are both odd positive integers, with gcd(s,t) = 1 and either
> > > > > t<s or t>3s.
>
> > > > > Class #2:
>
> > > > >    v = ((3s^2 + t^2)/4)^2
> > > > >    z = (s^4 - 10s^2 t^2 + 9 t^4)/16
>
> > > > > where s,t are both odd positive integers, with gcd(s,t) = 1 and either
> > > > > s>t or s<3t.
>
> > > > > I haven't carefully checked the algebra -- hopefully I haven't made
> > > > > errors. However I'm fairly sure the basic idea is correct, and I'm
> > > > > absolutely sure that there really are infinitely many primitive
> > > > > solutions.
>
> > > > > quasi
>
> > > > ...ah but if you mistyped the original equations into Magma in exactly
> > > > the same way that I did, you could have been as incorrect as I was! I
> > > > think
> > > > I did the "same" problem with v^2+3z^2 and v^2-z^2.....
>
> > > > You are of course correct. The question is one of pythagorean triples
> > > > (roughly) as you say...
>
> > > > (hanging head in shame) de P- Masquer le texte des messages précédents -
>
> > > > - Afficher le texte des messages précédents -
>
> > > Bonjour,
>
> > > Thanks for replying.
>
> > > I feel sorry , I've made a mistake :reading back
> > > my problem I've seen I expected three squares:
> > > v^2 , v^2-zv+2z^2 and v^2+3zv
> > > Okay your formula works for v^2+3zv a square,
>
> > > Best regards,
>
> > > Alain
>
> > Well, this problem really does lead to something of
> > genus 1 instead of 0! It corresponds to an elliptic
> > curve of rank 1 over Q, and conductor 1176 (again, this
> > is all assuming that I can type), so it seems that
> > there are infinitely many solutions of the desired shape
> > (such as v=529, z=440), which can be expressed recursively...
>
> > de P- Masquer le texte des messages précédents -
>
> > - Afficher le texte des messages précédents -
>
> Thanks a lot for your reply,
>
> Would you mind giving me some recursive expression,
> f.i from the given values v=529, z=440  ...
>
> Alain

I'm currently in a motel room, far from civilization. I think
this paper of Bremner

http://www.emis.de/journals/HOA/IJMMS/9/413.pdf

should lead you in the right direction (the problem is similar).

de P
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