A simple square equation
alainv...@gmail.com
(2^(p+1))^2 +(2^(2p+1) -1)^2 = 2^(2p+1) + 1
p =0 2 ^2 + 1^2 = 2^2 + 1^2
p =1 4^2 + 7^2 = 8^2 + 1^2
........................................................
p =5 64^2 + 2047^2 = 2048^2 + 1^2
Alain
William Elliot
> Form x^2 +y^2 = t^2 +1 , with non negative integer numbers.
>
> (2^(p+1))^2 + (2^(2p+1) - 1)^2 = 2^(2p+1) + 1
That's not of the form you indicated and it's false.
2^(2p + 2) + 2^(4p + 2) - 2*2^(2p+1) + 1 = 2^(2p+1) + 1
2^(4p + 2) = 2^(2p+1); 2^(2p + 1) = 1
alainv...@gmail.com
>
> - Afficher le texte des messages précédents -
Bonjour,
4^2+7^2 = 8^2+1 = 65
8^2+31^2 = 32^2+1= 1025
16^2+127^2=128^2+1 =13385
................
Yes,there was some miswriting
(2^(2p+2))^2 +(2^(2p+1) -1)^2 = (2^(2p+1))^2 + 1
Alain
Philippe 92
> On Tue, 5 Jan 2010, alainv...@gmail.com wrote:
>
>> Form x^2 +y^2 = t^2 +1 , with non negative integer numbers.
>>
>> (2^(p+1))^2 + (2^(2p+1) - 1)^2 = 2^(2p+1) + 1
>
> That's not of the form you indicated and it's false.
>
> 2^(2p + 2) + 2^(4p + 2) - 2*2^(2p+1) + 1 = 2^(2p+1) + 1
>
> 2^(4p + 2) = 2^(2p+1); 2^(2p + 1) = 1
>
>> p =0 2 ^2 + 1^2 = 2^2 + 1^2
>>
>> p =1 4^2 + 7^2 = 8^2 + 1^2
>
>> ...
Hi,
I didn't try to check your calculations, I suspect William is
right when He claims "and it's false" ;)
However for numbers x^2 + y^2 = t^2 + 1, you could as well use
the Fibonacci's formula for sum of squares.
It gives a generic set for all the x^2 + y^2 = t^2 + w^2, and
you can easily restrict it to w = 1 :
*********************************************************
* For any integers 0 < n < m, q > p > 0 and mn != pq *
* x = mp + nq, y = mq - np, t = mq + np, w = |mp - nq| *
* x^2 + y^2 = t^2 + w^2 *
*********************************************************
As you want w = |mp - nq| = 1, this gives p,q from m,n
using Euclides algo and Bï¿œzout relations.
(and therefore GCD(m,n) = 1)
Regards.
--
Philippe C., mail : chephip, with domain free.fr
site : http://mathafou.free.fr/ (mathematical recreations)
alainv...@gmail.com
> William Elliot wrote :
>
>
>
>
>
> > On Tue, 5 Jan 2010, alainvergh...@gmail.com wrote:
>
> >> Form x^2 +y^2 = t^2 +1 , with non negative integer numbers.
>
> >> (2^(p+1))^2 + (2^(2p+1) - 1)^2 = 2^(2p+1) + 1
>
> > That's not of the form you indicated and it's false.
>
> > 2^(2p + 2) + 2^(4p + 2) - 2*2^(2p+1) + 1 = 2^(2p+1) + 1
>
> > 2^(4p + 2) = 2^(2p+1); 2^(2p + 1) = 1
>
> >> p =0 2 ^2 + 1^2 = 2^2 + 1^2
>
> >> p =1 4^2 + 7^2 = 8^2 + 1^2
>
> >> ...
>
> Hi,
>
> I didn't try to check your calculations, I suspect William is
> right when He claims "and it's false" ;)
>
> However for numbers x^2 + y^2 = t^2 + 1, you could as well use
> the Fibonacci's formula for sum of squares.
> It gives a generic set for all the x^2 + y^2 = t^2 + w^2, and
> you can easily restrict it to w = 1 :
>
> *********************************************************
> * For any integers 0 < n < m, q > p > 0 and mn != pq *
> * x = mp + nq, y = mq - np, t = mq + np, w = |mp - nq| *
> * x^2 + y^2 = t^2 + w^2 *
> *********************************************************
>
> As you want w = |mp - nq| = 1, this gives p,q from m,n
>
> - Afficher le texte des messages précédents -
Dear Philippe,
What is wrong about?
4^2+7^2 = 8^2+1 = 65
8^2+31^2 = 32^2+1= 1025
16^2+127^2=128^2+1 =13385
Or
(2^(p+1))^2 +(2^(2p+1) -1)^2 = (2^(2p+1))^2 + 1
Alain
Philippe 92
> On 5 jan, 13:32, "Philippe 92" wrote:
>> William Elliot wrote :
>>> On Tue, 5 Jan 2010, alainvergh...@gmail.com wrote:
>>>> (2^(p+1))^2 + (2^(2p+1) - 1)^2 = 2^(2p+1) + 1
>>> That's not of the form you indicated and it's false.
>>
>> right when He claims "and it's false" ;)
>
>
> What is wrong about?
> (2^(p+1))^2 +(2^(2p+1) -1)^2 = (2^(2p+1))^2 + 1
>
Nothing now.
Because while I was posting, you corrected your typo
from "= 2^(2p+1) + 1" to "= (2^(2p+1))^2 + 1"
However this family of solutions to x^2 + y^2 = t^2 + 1^2 is valid,
although there are more general families, as I mentionned.
alainv...@gmail.com
> alainvergh...@gmail.com a crit :
>
>
>
>
>
> > On 5 jan, 13:32, "Philippe 92" wrote:
> >> William Elliot wrote :
> >>> On Tue, 5 Jan 2010, alainvergh...@gmail.com wrote:
> >>>> (2^(p+1))^2 + (2^(2p+1) - 1)^2 = 2^(2p+1) + 1
> >>> That's not of the form you indicated and it's false.
>
> >> I didn't try to check your calculations, I suspect William is
> >> right when He claims "and it's false" ;)
>
> > Dear Philippe,
>
> > What is wrong about?
> > ...
> > (2^(p+1))^2 +(2^(2p+1) -1)^2 = (2^(2p+1))^2 + 1
>
> Nothing now.
>
> Because while I was posting, you corrected your typo
> from "= 2^(2p+1) + 1" to "= (2^(2p+1))^2 + 1"
>
> However this family of solutions to x^2 + y^2 = t^2 + 1^2 is valid,
> although there are more general families, as I mentionned.
>
> Regards.
>
> --
> Philippe C., mail : chephip, with domain free.fr
>
> - Afficher le texte des messages précédents -
Dear Philippe,
Would you mind giving me an example of a 'building'
with you method?
Alain
Marc Bogaerts
x^2 +y^2 = t^2 +1 is equivalent to
x^2-t^2 = 1-y^2
or also
(t-x)*(t+x)=(y-1)*(y+1)
So if one can solve a*b=(y-1)*(y+1) for integers then
t=(a+b)/2, x=(b-a)/2 are good candidates.
Take for example 2*12=4*6 then t=7, x=5 and y:=5 and we get
5^2+5^2=7^2+1
but following example doesn't work:
3*8=6*8 gives t=19/2, x=13/2 and y=7 giving "half"-integers instead of
integers.
There are two cases depending on y being even or odd.
First case y=1 mod 2 so y=2*c+1
-------------------------------
The equation becomes
a*b=4*c*(c+1);
Another point is that in order to have no half-integer solutions we must
have a+b = 0 mod 2 or, equivalently a=b mod 2 which means a and b have
to be both even or both be odd. If they are both odd their product can
never give a multiple of 4 so the both have to be even, lets say a=2*d
and b=2*e;
Now the equation becomes:
d*e=c*(c+1)
giving t=d+e, x=e-d and y=2*c+1
Since we are the fifth today we start with c=5; so y=11;
the divisors of 30 are 1,2,3,5,6,10 15 and 30 we get following
decompositions for d*e:
1*30 giving 29^2+11^2=31^2+1
2*15 giving 13^2+11^2=17^2+1
3*10 giving 7^2 +11^2=13^2+1
Tomorrow we will have c=6 so y=13;
The divisors of 42 are 1,2,3,6,7,14,21 and 42 we get following
decompositions for d*e:
1*42 giving 41^2+13^2=43^2+1
2*21 giving 19^2+13^2=23^2+1
3*14 giving 11^2+13^2=17^2+1
Second case y is odd so y=2*c+1
-------------------------------
Then the equation becomes a*b=4*c-1
Now both a and b have to be odd, and, moreover
if a = 1 mod 4 then b = -1 mod 4 and vice versa.
But then a+b=0 mod 4 and so a=b mod 2 and there are no half-integer
solutions.
The case in your post corresponds to values where c is a power of two
(2^p) ,a=1 and a*b=b=4c^2-1. Then t=(a+b)/2= 2*2^p and
x=(b-a)/2=2*2^p= 2^(p+1)
If p is even then 4c^2-1= is divisible by 3 so one can take a=3 and
obtain other results.
Examples: p=4 then c=2^4=16 and y=32, and we can take a=3 and b=341 so that
t=172, and x=169 giving
169^2+32^2=172^2+1
This gives another family of "parametrized" solutions.
Philippe 92
> On 5 jan, 17:09, "Philippe 92" wrote:
>> alainvergh...@gmail.com wrote :
>>> (2^(p+1))^2 +(2^(2p+1) -1)^2 = (2^(2p+1))^2 ￜ+ 1
>>
>> However this family of solutions to x^2 + y^2 = t^2 + 1^2 is valid,
>> although there are more general families, as I mentionned.
>>
>
>
> Would you mind giving me an example of a 'building'
> with you method?
>
>>
>> * For any integers 0 < n < m, q > p > 0 and mn != pq *
>> * x = mp + nq, y = mq - np, t = mq + np, w = |mp - nq| *
>> *********************************************************
Choose any two m, n coprime, for instance m = 7, n = 3
Use Euclides algo to get p and q with pm - qn = 1
that is p = 1 + 3k, q = 2 + 7k
x = mp + nq = 7(1 + 3k) + 3(2 + 7k) = 13 + 42k
y = mq - np = 7(2 + 7k) - 3(1 + 3k) = 11 + 40k
t = mq + np = 7(2 + 7k) + 3(1 + 3k) = 17 + 58k
That is
*****************************************************
* (13 + 42k)^2 + (11 + 40k)^2 = (17 + 58k)^2 + 1 *
*****************************************************
x^2 + y^2 = (169 + 1092k + 1764k^2) + (121 + 880k + 1600k^2)
= 290 + 1972k + 3364k^2
t^2 = 289 + 1972k + 3364k^2
This results into the family :
13^2 + 11^2 = 17^2 + 1
55^2 + 51^2 = 75^2 + 1
97^2 + 91^2 = 133^2 + 1
139^2 + 131^2 = 191^2 + 1
...
You can generate as many such families as you choose m, n.
The (p, q) from (m, n) may be written as a closed form from
congruences and little Fermat theorem, but the resulting closed
form formula in m, n, k is IMHO unpracticable.
Just restrict to given (m, n) and one parameter k as above.
alainv...@gmail.com
> alainvergh...@gmail.com wrote :> >>> (2^(p+1))^2 +(2^(2p+1) -1)^2 = (2^(2p+1))^2 + 1
>
> - Afficher le texte des messages précédents -
Dear Philippe, Dear Marc,
Thanks a lot for your clear answers.
Moreover, we may find two different 'approaches'
to soln,
Alain
bill
Here's a way to generate integer x,y,t triplets
Assign arbitrary values to g & k.
If k is odd, then
x = g*k +1
m = (x-1)/k = ((g*k+1)-1)/k = g.
n = k*(x+1)
y = (n-m)/2
t = (n+m)/2
If k is even, then
x = 2*g*k+1
m = (x-1)/k = ((2*g*k+1)-1)/k = 2*g.
n = k*(x+1)
y = (n-m)/2
t = (n+m)/2
regards, Bill J
bill
If k & g are both even, then
x = g*k+1 or 2*g*k+1 NOTE: If x= g*k+1, then m = g;
m = g or 2*g. if x = 2*g*k+1, then m = 2*g.
Enrico
>
> - Show quoted text -
=====================================================
For any integers A, B, C
(A+B)^2 + (B+C)^2 = (A+B+C)^2 + R
R = B^2 - 2AC
For R = 1, choose any odd B.
A, C are the factors of (B^2-1) / 2
Enrico
Andrew Usher
> For any integers A, B, C
>
> (A+B)^2 + (B+C)^2 = (A+B+C)^2 + R
>
> R = B^2 - 2AC
>
> For R = 1, choose any odd B.
> A, C are the factors of (B^2-1) / 2
And it's easy to see that this gives all the solutions. But there's
another way, which is to use number theory, that must give the same
result. Weird.
Andrew Usher
alainv...@gmail.com
Bonjour,
I've read your clear propositions,
can't we sum up it all this way:
x^2+y^2 = t^2 +n^2 , n given.
or (y-n)*(y+n) =(t-x)*(t+x)
with x,y,t,p all integer numbers
(t-x)=(y-n)/p ; (t+x)=p(y+n)
these two last terms must have the same parity.
For n=1 :
Example y =7 , p =3
(t-x) =(7-1)/3 ;(t+x) = 3*(7+1)
So t=13, x=14 and 11^2 +7^2=13^2+1^2
.......
Nota:
There is a particular case with powers of 2
x=2^(q+1) , y = 2^(2q+1)-1,t =2^(2q+1) ,
Example:
q=3 , 16^2+127^2 = 128^2 +1
Best regards,
Alain
Andrew Usher
wrote:
> I've read your clear propositions,
> can't we sum up it all this way:
>
> x^2+y^2 = t^2 +n^2 , n given.
> or (y-n)*(y+n) =(t-x)*(t+x)
> with x,y,t,p all integer numbers
> (t-x)=(y-n)/p ; (t+x)=p(y+n)
> these two last terms must have the same parity.
Yes, this is another way. But I prefer Enrico's which can be
generalised even more: set x,y,z = his A+B,B+C,A+B+C - this is one-to-
one as the latter vectors are lin. indep. - and we have
x^2 + y^2 = z^2 + n
is solved by seeing that n = (x+y-z)^2 - 2*(z-x)*(z-y) and going
through the possibilities for x,y,z. This is more general as the
constant term need not be a square.
The number-theoretical solution I spoke of is that z^2 + 1 , being the
sum of two coprime squares, must be (if odd) a product of primes 1 mod
4 or (if even) double of such a number. The number of ways to write
such a number as the sum of coprime squares is 2^k where k is the
number of odd prime factors - the number of ways as the sum of non-
coprime squares is easily derived from that. Exactly when that number
is greater than 1, there will be a non-trivial solution of x^2 + y^2 =
z^2 + 1 .
Andrew Usher