Interesting number theory-related problem.

[Mathguy15]

I have almost no idea how to start! I'm just doing this problem for fun(or was at one time):

Find all a such that the equation x^2+axy+y^2=1 has infinitely many solutions in integers for x and y. I know that the solution of this problem would follow from a complete understanding of integer squares of the form (b^2-4)(c^2-4) for integers b and c. If I could achieve an understanding of expressions of this form, I could solve the problem. What are your thoughts on this problem?

 

[epsi00]

start with a numerical example of your choice ( for the value of a ) and use http://www.alpertron.com.ar/QUAD.HTM
to solve "step by step". See what comes thru.

 

[Mathguy15]

start with a numerical example of your choice ( for the value of a ) and use http://www.alpertron.com.ar/QUAD.HTM
to solve "step by step". See what comes thru.

I will check it out. Thanks!

 

[robert2734]

There is a rotation by an angle theta that will make the axy term disappear. You have a conic section, perhaps degenerate, that is rotated with respect to the x and y axis. make the substitution x=x'cos(t)+y'sin(t) y=-x'sin(t)+y'cos(t) and find angle t that makes the crossterm disappear.

 

[Mathguy15]

There is a rotation by an angle theta that will make the axy term disappear. You have a conic section, perhaps degenerate, that is rotated with respect to the x and y axis. make the substitution x=x'cos(t)+y'sin(t) y=-x'sin(t)+y'cos(t) and find angle t that makes the crossterm disappear.

Sorry, I don't know calculus or much trigonometry yet. I will soon, but not yet.

 

[robert2734]

Since the problem is symmetric in x and y, the rotation we want is 45 degrees and the substitution is x=sqrt(2)/2[u+v] y=sqrt(2)/2[u-v].

x^2=.5(u^2+2uv+v^2) axy=.5a(u^2-v^2) y^2=.5(u^2-2uv+v^2)

combining terms we have (1+a/2)u^2+(1-a/2)v^2=1.

There is no uv crossterm so we have accomplished our mission. We can see this equation is an ellipse if |a|<2. A hyperbola if |a|>2, and two parallel lines if |a|=2.

Anytime the coefficient of the x^2 and y^2 term is the same, the substitution x=u+v y=u-v will clear the crossterm.

 

[dodo]

Another possible way of getting rid of the cross term is by completing the square:\begin{align*}<br /> x^2 + axy + y^2<br /> &amp; = x^2 + axy + ( \frac a 2 )^2 y^2 - ( \frac a 2 )^2 y^2 + y^2 \\<br /> &amp;= (x + \frac a 2 y)^2 + (1 - \frac {a^2} 4) y^2<br /> \end{align*}
Substituting u = x + a/2 y, your original equation becomes<br /> u^2 + (1 - \frac {a^2} 4) y^2 = 1<br />
and, once you have a solution for u and y, you can recover a value for x.

My eyeball guess is that values of |a| < 2, making the last equation an ellipse, won't have a chance to result in infinite integer solutions. Which more or less coincides with what Robert said.

 

[robert2734]

A hyperbola won't have infinite integer solutions either. Eventually the curve is within one unit of the asymptote and no more integer solutions are possible. So the answer to your original problem is a=2 and a=-2.

 

[dodo]

Hmm... Robert, something must be wrong here. In the equationu^2 + (1 - \frac {a^2} 4) y^2 = 1if you take a=4 you get the equation u^2 - 3 y^2 = 1, which has infinite integer solutions (namely, u=numerator and y=denominator of every even-indexed convergent, \frac {p_2}{q_2}, \frac {p_4}{q_4}, \frac {p_6}{q_6} ... of the continued fraction for sqrt(3)).

(In this example, since a=4 and the substitution x = u - a/2 y was used, integers values for u and y produce also integer values for x.)

P.S:
More clearly, with a=4 the original equation is x^2 + 4xy + y^2 = 1, with integer solutions
(0,1), (-1,4), (-4,15), (-15,56), (-56,209), (-209,780), ...

P.P.S:
I think the issue is that, of all the points on the asymptote (a line which is neither horizontal nor vertical, if the hyperbola is on the standard axes) an infinitude of them will have non-integer coordinates. So there may easily be points on the asymptote which are very close to points with integer coordinates. *** Correction (sorry): there is a cross term, so the hyperbola was certainly NOT on the standard axes. But anyway, the asymptotes are not necessarily horizontal or vertical.

 

[Matt Benesi]

For integer values of a you get the following patterns (formula follows):

a=1 : (0,1) (-1,1) (-1,1) ... stays the same so it's excluded.

a=2 : (0,1) (-1,2) (-2,3) (-3,4) (-4,5) ...

Basically the counting numbers, with reversed signs. Note that if a is negative, x and y should be the same sign, if a is positive, x and y have to be the opposite sign.

a=3 : Just take 2 consecutive numbers http://oeis.org/A001906" as x and y and you'll get 1.

a=4 : http://oeis.org/A001353" ...

Basically the formula for integer x and y that work for integer |a|>1 is

f_n= a * f_n-1 - f_n-2

Start out with f_n-2=0 and f_n-1=1

So for a=9 f_n= 9*1-0 = 9

next value would be: 9*9 -1 = 80

next: 9* 80 -9 = 711... so on and so forth (http://oeis.org/A018913" ).


The best way to think of a non-integer a is as a fraction: a= b/c

In your series of generated numbers that will give you x and y, you'll continue to have powers of c in the denominator, although in certain instances they might cancel out with the numerator? Anyone?.

So non-integer a's can produce non-integer patterns (such as an a of 2.5):

5/2 *1 - 0 = 5/2
5/2 *5/2 - 2/2 = 25/4 -4/4 = 21/4
21/4 *5/2 - 5/2 =105/8 - 20/8 = 85/8...

The numerator will always end in a 1 or 5 (it's http://oeis.org/A002450" ]), so will never be divisible by 2, and never cancel out.

 

[Matt Benesi]

Here is the generating function for series of numbers that work for a>2 (a=2 is simply consecutive integers with appropriate signs).

Set a to some integer value (or whatever if you want to mess around). The value n is the depth into the series (so when n=2 you get the second number of the series, n=0 is 0, n=1 is 1, etc...).

h\left( a,n\right) :=\frac{{\left( \frac{a+\sqrt{{a}^{2}-4}}{2}\right) }^{n}-{\left( \frac{a-\sqrt{{a}^{2}-4}}{2}\right) }^{n}}{\sqrt{{a}^{2}-4}}​