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A Quartic Diophantine Equation: 10657 + 11579x^2 + x^4 = y^2

Date: 12/29/2008 at 15:16:31
From: colin
Subject: The Diophantine equation 10657 + 11579x^2 + x^4 = y^2

How can you find at least one solution to this Diophantine equation?

  10657 + 11579x^2 + x^4 = y^2

Why is it solvable?  or is it not?  What is the general method for 
solving these kind of equations?

I have searched the web and found a lot of suggestions, including on 
this site.  I even found a question about nearly the same equation.  
Unfortunately, if you look at the general equation ax^4 + bx^2 + c, 
the c value is a perfect square most of the time. That has nothing to 
do with this equation, since x = 0 always returns a perfect square if 
c is a perfect square.

My web searches also led me to a document from Tzanakis in which he 
solves a general quartic equation by expanding it in some way to an 
equation with a quadratic constant factor, c.  I do not understand 
how he does this.  Nor do I understand the use of this and how to 
convert the results back to the original equation.  He also assumes 
that the equation has at least one solution.  In general, the math in 
that document goes way beyond the math I understand. 

My guess is that it must somehow be possible to determine whether 
this equation has solutions at all.  After that, the next problem 
becomes converting it into an elliptic curve of Weierstrass form.  I 
do not have enough knowledge to do that since most of the quartic 
polynomials I found are special form polynomials which can be easily 
converted into an elliptic curve.



Date: 07/01/2009 at 17:01:18
From: Doctor Vogler
Subject: Re: The Diophantine equation 10657 + 11579x^2 + x^4 = y^2

Hi Colin,

Thanks for writing to Dr. Math.  I apologize for the huge delay; we 
don't normally answer questions after they are this old, but I love 
Diophantine equations, and I noticed this question in the archive.  I 
must have missed it over Christmas.

Because the leading coefficient of the quartic is a square (or 
rather, the ratio of the leading terms of each side is a square), you 
can rewrite your equation as a difference of squares, like so:

  (2x^2 + 11579)^2 - (2y)^2 = 134030613

Then you can factor it

  (2x^2 + 11579 + 2y)(2x^2 + 11579 - 2y) = 134030613

From here, use the ideas a colleague posted in this Dr. Math 
conversation:

  Second-Degree Two-Variable Diophantine Equation
    http://mathforum.org/library/drmath/view/55988.html 

Namely, since you want x and y to be integers, the left side must be 
a product of two integers; and therefore it gives a factorization of 
the integer on the right side, 134030613.  This number factors as

  134030613 = 3 * 37 * 43 * 28081

Now look at the concept presented in another Dr. Math exchange:

  How Many Factors?
    http://mathforum.org/library/drmath/view/54227.html 

The 32 integer factors of 134030613 are ...

  1
  3
  37
  43
  3*37 = 111
  3*43 = 129
  37*43 = 1591
  3*37*43 = 4773
  28081
  3*28081 = 84243
  37*28081 = 1038997
  43*28081 = 1207483
  3*37*28081 = 3116991
  3*43*28081 = 3622449
  37*43*28081 = 44676871
  3*37*43*28081 = 134030613

... and the negatives of those numbers.

We have

  (2x^2 + 11579 + 2y)(2x^2 + 11579 - 2y) = 134030613

That means that ...

  2x^2 + 11579 + 2y

... must be one of those 32 factors.  So we can try each of them in 
turn.  For example, suppose that

  2x^2 + 11579 + 2y = 28081.

Then that means that

  2x^2 + 11579 - 2y = 134030613/(2x^2 + 11579 + 2y)
                    = 134030613/28081
                    = 4773

So their sum is

  4x^2 + 23158 = 28081 + 4773

We can solve this for x^2 as

  x^2 = 2424

This not a perfect square (not the square of an integer).

If you try all 32 factors, then you can find any integer solutions to 
your Diophantine equation.  (In fact, it doesn't take much thought to 
rule out the 16 negative factors, so you can really just check the 16 
positive ones.  But try at least one negative factor to see why you 
can eliminate all of them.)

It turns out that you don't get a perfect square for x^2 using any of 
the 32 factors.  So there are no integer solutions.

If you have any questions about this or need more help, please write 
back and show me what you have been able to do, and I will try to 
offer further suggestions.

- Doctor Vogler, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 07/03/2009 at 16:56:34
From: colin
Subject: Thank you (The Diophantine equation 10657 + 11579x^2 + x^4 = 
y^2)

Thank you very much for your extensive answer.

Sincerely,

Colin
Associated Topics:
High School Linear Equations
High School Number Theory

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