The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Solving a Diophantine Equation By Use of Number Fields

Date: 03/01/2006 at 03:36:50
From: Krassi
Subject: x^2 = y^7 + 7

Prove that there aren't integer solutions (x,y) to the equation:
x^2 = y^7 + 7.  It seems like x must be even and y must be 1(mod 4),
but I'm having trouble getting anywhere.

Date: 03/01/2006 at 21:32:18
From: Doctor Vogler
Subject: Re: x^2 = y^7 + 7

Hi Krassi,

Thanks for writing to Dr. Math.  The easy way to answer questions of
this type is to find some modulus m for which

  x^2 = y^7 + 7 (mod m)

has no solutions, because then you can check them all to verify that
there are no solutions to this congruence, and that implies that there
are no integer solutions.

But sometimes there are no integer solutions, and yet there are
solutions to that congruence for every m.

So my first thought was to find a prime number m = p such that the
multiplicative group mod p has order (p-1) which is a small multiple
of 7.  Since

  p - 1 = 14

gives a non-prime, we try

  p - 1 = 28.

Then there are only four 7th powers mod 29.  That's a good sign!  If
you add 7 to any of them, do you ever get a square mod 29?

Unfortunately, it turns out that you do.  You can try higher primes,
but the higher you go, the less likely it is going to work.

There is another method which is more likely to work, but it requires
some more sophisticated math.  It involves number fields.  I talked
about number fields in the second half of the answer

  Solving with the Pell Equation 

which is perhaps worth reading.  Without going into all of the
details, you can solve your equation by factoring

  y^7 = x^2 - 7


  y^7 = (x + sqrt(7))(x - sqrt(7)),

and then considering how numbers factor in the number field
Q(sqrt(7)).  Just as if you had an equation like

  y^7 = (x + 2)(x - 3),

you would want to see if the two factors on the right are relatively
prime and, if they are, that means that each one must be a seventh
power.  In our case, the gcd of x + sqrt(7) and x - sqrt(7) must be a
factor of their difference

  2*sqrt(7) = (3 - sqrt(7))*(3 + sqrt(7))*sqrt(7),

which is a product of three primes as written.  (The primes in a
number field are not always the same as normal or "rational" primes,
but 3 - sqrt(7) and 3 + sqrt(7) and sqrt(7) are all prime in our
number field.)

Well, it turns out that x has to be odd for either 3 + sqrt(7) or 3 -
sqrt(7) to divide x + sqrt(7).  Since you already determined that x is
even, we can rule out that possibility.  Furthermore, x must be
divisible by 7 for sqrt(7) to divide x + sqrt(7).  But if x is
divisible by 7, then so is y, but this yields a contradiction in the
original equation (why?).

So we conclude that x + sqrt(7) and x - sqrt(7) are relatively prime.
Therefore, each is a seventh power,

  x + sqrt(7) = (a + b*sqrt(7))^7
          = (a^7 + 147*b^2*a^5 + 1715*b^4*a^3 + 2401*b^6*a) + 
            (7*b*a^6 + 245*b^3*a^4 + 1029*b^5*a^2 + 343*b^7)*sqrt(7)
          = (a)(a^6 + 147*b^2*a^4 + 1715*b^4*a^2 + 2401*b^6) + 
            (7b)(a^6 + 35*b^2*a^4 + 147*b^4*a^2 + 49*b^6)*sqrt(7).

But this says that the coefficient of sqrt(7) on the left side is 1,
but the coefficient on the right side is a multiple of 7.  Since 1
isn't a multiple of 7, that means that there cannot be an integer
solution to your equation.

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 
Associated Topics:
College Number Theory

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.