My Math Forum Integer solutions to (ax^2+bx+c)+(ay^2+by+c) = (az^2+bz+c)

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 January 8th, 2017, 07:49 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, 08:14 PM #2 Member   Joined: Jul 2015 From: tbilisi Posts: 30 Thanks: 8 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. Thanks from Naveenchandra Kumar
 January 8th, 2017, 11:11 PM #3 Member   Joined: Jul 2015 From: tbilisi Posts: 30 Thanks: 8 For Diophantine equation $\displaystyle aX^2+bX+c+aY^2+bY+c=aZ^2+bZ+c$ If the root is integer. $\displaystyle q=\sqrt{b^2-4ac}$ We use the solutions of the equation Pell. $\displaystyle p^2-2(k-s)ks^2=1$ The solution then can be written in this form. $\displaystyle X=\frac{1}{a}(\frac{-b\pm{q}}{2}p^2+(k-s)qps+(b\pm{q})(k-s)ks^2)$ $\displaystyle Y=\frac{1}{a}(\frac{-b\pm{q}}{2}p^2+kqps+(b\pm{q})(k-s)ks^2)$ $\displaystyle Z=\frac{1}{a}(\frac{-b\pm{q}}{2}p^2+(2k-s)qps+(b\pm{q})(k-s)ks^2)$ To find all solutions is necessary to solve the more General equation. With different coefficients. Thanks from Naveenchandra Kumar
January 9th, 2017, 04:48 PM   #4
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I am trying to understand the answer given by individ

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 Originally Posted by Individ For Diophantine equation $\displaystyle aX^2+bX+c+aY^2+bY+c=aZ^2+bZ+c$ If the root is integer. $\displaystyle q=\sqrt{b^2-4ac}$ We use the solutions of the equation Pell. $\displaystyle p^2-2(k-s)ks^2=1$ The solution then can be written in this form. $\displaystyle X=\frac{1}{a}(\frac{-b\pm{q}}{2}p^2+(k-s)qps+(b\pm{q})(k-s)ks^2)$ $\displaystyle Y=\frac{1}{a}(\frac{-b\pm{q}}{2}p^2+kqps+(b\pm{q})(k-s)ks^2)$ $\displaystyle Z=\frac{1}{a}(\frac{-b\pm{q}}{2}p^2+(2k-s)qps+(b\pm{q})(k-s)ks^2)$ To find all solutions is necessary to solve the more General equation. With different coefficients.

 January 9th, 2017, 09:55 PM #5 Member   Joined: Jul 2015 From: tbilisi Posts: 30 Thanks: 8 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. $\displaystyle k=5$ ; $\displaystyle s=3$ $\displaystyle p^2-20s^2=1$ 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. Thanks from Naveenchandra Kumar

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