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From: alainverghote on 8 Aug 2010 06:54
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
From: The Pumpster on 8 Aug 2010 12:55
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.

de P
From: quasi on 8 Aug 2010 16:43
On Sun, 8 Aug 2010 09:55:52 -0700 (PDT), The Pumpster
<pumpledumplekins(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
From: The Pumpster on 8 Aug 2010 18:47
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
From: alainverghote on 9 Aug 2010 06:32
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
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