# solution of equations

Max Alekseyev maxale at gmail.com
Sun Mar 18 02:50:28 CET 2007

```On 3/17/07, franktaw at netscape.net <franktaw at netscape.net> wrote:> I just received the following, which I am forwarding to the list> without further comment.
I've also got such email. This was my reply:
-------- Original Message --------Subject: Re: solution of equationsDate: Sat, 17 Mar 2007 17:57:19 -0700To: Mohamed Bouhamida <bhmd95 at yahoo.fr>
Mohamed Bouhamida wrote:> *                                                   March 17, 2007 *> Dear Colleague;> I have the honor to address to you to inform you that I am an Algerian researcher in Mathematics, born in 1969 at Ain-sefra (- w-Naâma) and that after a research which lasted more than ten (10) years I succeeded in solving some problems which one can pose in natural numbers.
These equations are easy and/or well-studied. You did not need tospent 10 years proving them.See below.
> I wanted to share with you the results of my research.> *1/* *Solution of the equation:X2+(X+K)2=Y2 with K is a prime * number and *K=2*m2-1, m>=2*> * ex *: K=7,17,31,71,...> the X values are given by the sequence defined by: a(n)=6*a(n-3)-a(n-6)+2K with:> a(0)=0,   a(1)=2m+1,               a(2)=6*m2-10m+4> a(3)=3K, a(4)=6*m2+10m+4, a(5)=40*m2-58m+21.> click here: http://www.research.att.com/~njas/sequences/A118674 <http://www.research.att.com/~njas/sequences/A118674>                                          *2/* *Solution of the equation:X2+(X+K)2=Y2 with K is a prime * number and  *K=p2-2, p>=5*> * ex *: K=23,47,79,...> the X values are given by the sequence defined by: a(n)=6*a(n-3)-a(n-6)+2K with:> a(0)=0,   a(1)=2p+2,               a(2)=3*p2-10p+8> a(3)=3K,  a(4)=3*p2+10p+8, a(5)=20*p2-58p+42.> click here: http://www.research.att.com/~njas/sequences/A118675 <http://www.research.att.com/~njas/sequences/A118675>
X2+(X+K)2=Y2 in equivalent form is2*X2 + 2*K*X - Y2 + K2 = 0.
This is a partial case of well-studied generalized Pell equations.There are tons of literature on them.Moreover, there is an on-line tool that solves such equations:http://www.alpertron.com.ar/QUAD.HTMBy request, it can even produce step-by-step solutions.
Using this tool, your solutions can be obtained in a matter of minutes.
> **  * 3/ Solution of the equation:(X-Y)3-XY=0 *> the X values are given by the sequence defined by: a(n)=n*(n+1)2, with n is a natural integer.> click here: http://www.research.att.com/~njas/sequences/A045991 <http://www.research.att.com/~njas/sequences/A045991>> the Y values are given by the sequence defined by: b(n)=(n+1)*n2, with n is a natural integer.> click here: http://www.research.att.com/~njas/sequences/A011379 <http://www.research.att.com/~njas/sequences/A011379>> **  * 4/T he equation:(X-Y)^m-XY=0, with m is an even number,  m>=2 *> I proved that this type of equations has no solution of course without taking the solution (0,0) in consideration.click here: http://www.research.att.com/~njas/sequences/A045991>> * 5/T he equation:(X-Y)^m-XY=0, with m is an odd number, m>=5 *> I could not until now find solutions,also I could not prove that this type of equations has no solution.> Best regards;> Mohamed Bouhamida.( bhmd95 at yahoo.fr <http://fr.f270.mail.yahoo.com/ym/Compose?To=bhmd95@yahoo.fr>)
Consider equation (X-Y)^m=XY. Let k=X-Y be an non-negative integer.Then the original equation is equivalent to two:X-Y=k and XY=k^m.We write the latter one as X*(-Y)=-k^m.By Vieta's formulas, we have that X and (-Y) are the roots of thefollowing quadratic equation:z2 - k*z - k^m = 0.Then its discriminant k2 + 4*k^m = k2*(1 + 4*k^(m-2)) must a square,and so is 1 + 4*k^(m-2).Hence, we need to solve the following equation in non-negative integers:1 + 4*k^(m-2) = t2
It is easy to see that for even m>=2, there is no solutions except k=0and t=1. This proves your case 4/ above.
For m=3, we have1 + 4*k = t2implying that t=2*n+1 and k=n2+n for any non-negative integer n.That givesX = (k + k*t)/2 = (n+1)2*nY = (k*t - k)/2 = (n+1)*n2That indeed match your solutions in the case 3/ above.
Finally, for odd m>=5, we can put our equation as4*k^(m-2) = t2 - 1 = (t-1)*(t+1)Then since gcd(t-1,t+1)=2 we have t-1 = 2*u^(m-2) and t+1 = 2*v^(m-2) and, thus,v^(m-2) - u^(m-2) = 1.Now we can use Catalan conjecture (that was proved in 2002) statingthat the only perfect powers differing in 1 are 32 - 23 = 1.Catalan conjecture gives that the equation v^(m-2) - u^(m-2) = 1 hasno solutions.Therefore, in your case 5/ there are no solutions.
Regards,Max

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