Diophantine equation second grade

[Lolyta]

Hello!
I am trying to find the general solution of this equation, could you help me?
\(\displaystyle \dfrac{x(x-1)}{{y(y-1)}} =1/2\)

Thank you so much!

 

[Deleted member 4993]

Hello!
I am trying to find the general solution of this equation, could you help me?
\(\displaystyle \dfrac{x(x-1)}{{y(y-1)}} =1/2\)

Thank you so much!

By observation -particular solution:

\(\displaystyle \dfrac{3*2}{4*3} \ = \ \dfrac{1}{2}\)

What can you deduce from that?

 

[soroban]

Hello, Lolyta!

\(\displaystyle \text{Find the general solution of: }\:\dfrac{x(x-1)}{{y(y-1)}} \:=\:\dfrac{1}{2}\)

We have: .\(\displaystyle 2x(x-1) \:=\:y(y-1) \quad\Rightarrow\quad 2(x^2 - x) \:=\:y^2-y \)


Complete the square: .\(\displaystyle 2(x^2 - x + \frac{1}{4}) - \frac{1}{2} \:=\: (y^2 - y + \frac{1}{4}) - \frac{1}{4}\)

. . . . . . . . . . . . . . \(\displaystyle 2(x-\frac{1}{2})^2 - (y - \frac{1}{2})^2 \;=\;\frac{1}{4}\)

. . . . . . . . . . . . . .\(\displaystyle 8(x-\frac{1}{2})^2 - 4(y - \frac{1}{2})^2 \;=\;1\)

. . . . . . . . . . . . . . .\(\displaystyle \dfrac{(x-\frac{1}{2})^2}{\frac{1}{8}} - \dfrac{(y-\frac{1}{2})^2}{\frac{1}{4}} \;=\;1\)


\(\displaystyle \text{The general solution is }(x,y)\text{, any point on this hyperbola.}\)

 

[Deleted member 4993]

Nice work Soroban - I did not think along that line.

The title line - "2 nd. grade" - had me confused enough so that I did not think of analysis.

The other integer solutions would be (-2,-3), (3,-3), (-2,4).

Are there other integer (Diophantine) solutions? ....Denis???

 

[daon2]

Pretty sure I didn't do these kinds of problems until high school. In fact I don't believe I saw exponents until 7th or 8th grade.

 

[Lolyta]

Hello, Lolyta!


We have: .\(\displaystyle 2x(x-1) \:=\:y(y-1) \quad\Rightarrow\quad 2(x^2 - x) \:=\:y^2-y \)


Complete the square: .\(\displaystyle 2(x^2 - x + \frac{1}{4}) - \frac{1}{2} \:=\: (y^2 - y + \frac{1}{4}) - \frac{1}{4}\)

. . . . . . . . . . . . . . \(\displaystyle 2(x-\frac{1}{2})^2 - (y - \frac{1}{2})^2 \;=\;\frac{1}{4}\)

. . . . . . . . . . . . . .\(\displaystyle 8(x-\frac{1}{2})^2 - 4(y - \frac{1}{2})^2 \;=\;1\)

. . . . . . . . . . . . . . .\(\displaystyle \dfrac{(x-\frac{1}{2})^2}{\frac{1}{8}} - \dfrac{(y-\frac{1}{2})^2}{\frac{1}{4}} \;=\;1\)


\(\displaystyle \text{The general solution is }(x,y)\text{, any point on this hyperbola.}\)

Thank you very much to everyone . It is fantastic!
But, in fact, I was looking for natural solutions (because it is an step for solving a particular exercise).
Thanks again!

 

[Lolyta]

Why pick on poor me?
3,-3
3,4
-2,-3
-2,4

Hi to everyone.
Wolfram alpha ( wolframalpha . com) has shown me the natural solutions:
Let \(\displaystyle a= (3-2\sqrt{2})^n\) and \(\displaystyle b= (3+2\sqrt{2})^n\)

\(\displaystyle x= \frac{1}{8}(2a+a\sqrt{2}+2b-b\sqrt{2}+4) ; y= \frac{1}{4}(-a-a\sqrt{2}-b+b\sqrt{2}+2) ; \forall{n>1} \) with n an integer
If we replace n for 2 we have (3,4) for 3 the next solution...
Do you have any idea of how has it been done? I mean, the "code" or algorithm?

Just in case it would be useful I put here the general solution: \(\displaystyle y= \frac{1}{2}(1+ \sqrt{8x^2-8x+1}) \) but it's for real numbers, I want just positive integers just how the form below

Thank you!

 

[Lolyta]

That's simply the result of solving your original equation : x(x-1) / y(y-1) = 1/2
in terms of y, using the quadratic equation.

Yes, I have put that just in case it was useful to find this result:
\(\displaystyle x= \frac{1}{8}(2a+a\sqrt{2}+2b-b\sqrt{2}+4) ; y= \frac{1}{4}(-a-a\sqrt{2}-b+b\sqrt{2}+2) ; \forall{n>1} \) with n an integer
Well, my question was if you know how could we reach that solution. This is what i need, just the positive integers.
Thanks for everything.