
Number Theory Number Theory Math Forum 
 LinkBack  Thread Tools  Display Modes 
January 8th, 2017, 08:49 PM  #1 
Newbie Joined: Jan 2017 From: Mangalore, Karnataka, India Posts: 11 Thanks: 0  Integer solutions to (ax^2+bx+c)+(ay^2+by+c) = (az^2+bz+c)
Is there any general method to find integer solutions to these equations?

January 8th, 2017, 09:14 PM  #2 
Newbie Joined: Jul 2015 From: tbilisi Posts: 26 Thanks: 7 
Yes  there is. To reduce this equation to a certain is equivalent to the Pell equation. And the solution of this equation will be determined by the solutions of the equation Pell. And for this particular case you can think of the parameterization and without equations Pell. 
January 9th, 2017, 12:11 AM  #3 
Newbie Joined: Jul 2015 From: tbilisi Posts: 26 Thanks: 7 
For Diophantine equation $\displaystyle aX^2+bX+c+aY^2+bY+c=aZ^2+bZ+c$ If the root is integer. $\displaystyle q=\sqrt{b^24ac}$ We use the solutions of the equation Pell. $\displaystyle p^22(ks)ks^2=1$ The solution then can be written in this form. $\displaystyle X=\frac{1}{a}(\frac{b\pm{q}}{2}p^2+(ks)qps+(b\pm{q})(ks)ks^2)$ $\displaystyle Y=\frac{1}{a}(\frac{b\pm{q}}{2}p^2+kqps+(b\pm{q})(ks)ks^2)$ $\displaystyle Z=\frac{1}{a}(\frac{b\pm{q}}{2}p^2+(2ks)qps+(b\pm{q})(ks)ks^2)$ To find all solutions is necessary to solve the more General equation. With different coefficients. 
January 9th, 2017, 05:48 PM  #4  
Newbie Joined: Jan 2017 From: Mangalore, Karnataka, India Posts: 11 Thanks: 0  I am trying to understand the answer given by individ Quote:
 
January 9th, 2017, 10:55 PM  #5 
Newbie Joined: Jul 2015 From: tbilisi Posts: 26 Thanks: 7 
There is a parable of the Asian philosopher Nasreddin. One fool can ask so many questions that they do not answer 100 scientists! You ask the question  the difficulty level you do not understand. You want to get a simple answer? It does not happen. Nature itself decides how the law should work, and how to look. We can only to be measured with this. For example, such factors. $\displaystyle q=5$ ; $\displaystyle a=3$ ; $\displaystyle b=11$ ; $\displaystyle c=8$ $\displaystyle 3X^2+11X+8+3Y^2+11Y+8=3Z^2+11Z+8$ Now you have to choose something a Pell. For example. $\displaystyle k=5$ ; $\displaystyle s=3$ $\displaystyle p^220s^2=1$ All the solutions of this equation can be found if we take any decision to substitute in the formula and obtain the following solution. $\displaystyle p_2=9p+40s$ $\displaystyle s_2=2p+9s$ We must begin with the first solution. $\displaystyle (p;s)  (9;2) $ Next will be such. $\displaystyle (p;s)  (161 ; 36) $ .......... Now substitute the coefficients in the formula. $\displaystyle X=\frac{1}{3}(3p^2\pm10ps+160s^2)$ $\displaystyle Y=\frac{1}{3}(3p^2\pm25ps+160s^2)$ $\displaystyle Z=\frac{1}{3}(3p^2\pm35ps+160s^2)$ Substitute such second solution. The difference arises because the sign that Pell's equation solutions can have any sign. $\displaystyle (p,s)  (\pm161 ; \pm36)$ Numerical values will be such. $\displaystyle X=62519$ $\displaystyle Y=91499$ $\displaystyle Z=110819$ And so on. $\displaystyle X=23879$ $\displaystyle Y=5101$ $\displaystyle Z=24421$ The point is simple. Find the solution to Pell's equation, then we substitute into the formula and get the right solution. 

Tags 
ax2, ay2, az2, integer, solutions 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Integer solutions to (x^2+x) + (y^2+y) = (z^2+z)  Naveenchandra Kumar  Algebra  4  January 9th, 2017 12:03 PM 
Integer solutions to (x^2 + x + 2) + (y^2 + y + 2) = (z^2 + z + 2)?  Naveenchandra Kumar  Number Theory  4  January 8th, 2017 09:10 PM 
Integer Solutions to ZēsZ+p=0  AlephUser  Elementary Math  0  January 3rd, 2014 06:42 AM 
all integer solutions of 3^m2^n=+1  calligraphy  Number Theory  7  April 13th, 2011 12:56 PM 
x^5+31=y^2 has no integer solutions  elim  Number Theory  4  April 5th, 2010 01:05 PM 