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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.

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)

Sincerely,

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

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