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Finding Integer Solutions of x^3 - y^2 = 2


Date: 06/01/2000 at 13:20:29
From: Carlos Henrique Griese Neto
Subject: Finding the integer solutions

Hi. My name is Carlos and I'm from Brazil. I'm trying to solve this 
problem, but I'm stuck. How can I find all integer solutions of the 
equation x^3 - y^2 = 2? Does the equation x^2 - y^3 = 2 have integer 
solutions?

Thanks a lot.


Date: 06/01/2000 at 16:54:39
From: Doctor Rob
Subject: Re: Finding the integer solutions

Thanks for writing to Ask Dr. Math, Carlos.

The only integer solutions of x^3 - y^2 = 2 are (3,5) and (3,-5). I am 
told that the proof of this is very hard.

x^2 - y^3 = 2 has the solutions (1,-1) and (-1,-1), at least.

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


Date: 06/02/2000 at 09:54:05
From: Doctor Rob
Subject: Re: Finding the integer solutions

Just a follow-up to my previous answer. I now have a proof that (3,5) 
and (3,-5) are the only integer solutions to x^3 - y^2 = 2. It uses 
arithmetic in the ring Z[sqrt(-2)]. If you need this proof, write 
back.

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


Date: 06/02/2000 at 16:40:55
From: Carlos H. Griese Neto
Subject: Re: Finding the integers solutions

Hi, it's me again.

Thanks for your answer.

I'd be very happy if you could send me the proof that the only integer 
solutions of the equation x^3 - y^2 = 2 are (3,5) and (3,-5).

Thanks again,
Carlos H.


Date: 06/02/2000 at 17:02:07
From: Doctor Rob
Subject: Re: Finding the integers solutions

Okay, here it is.

Z[sqrt(-2)] is a unique factorization domain, and its only units are 
1 and -1. Write

     x^2 = y^2 + 2 = (y + sqrt[-2])*(y - sqrt[-2])

Notice that GCD(y+sqrt[-2], y-sqrt[-2]) | 2*sqrt[-2]. Now it follows 
from unique factorization and the fact about the units that for some 
z in Z[sqrt(-2)],

     y + sqrt(-2) = z^3  or  z^3*sqrt(-2)  or  z^3*(-2)

Taking complex conjugates, we then have

     y - sqrt(-2) = w^3  or  w^3*sqrt(-2)  or  w^3*(-2)

where w is the conjugate of z. Hence, multiplying these together,

     x^3 = y^2 + 2 = (z*w)^3  or  -2*(z*w)^3  or  4*(z*w)^3

Since w*z is in Z, unique factorization in Z implies that

     y^2 + 2 = (z*w)^3,

and

     y + sqrt(-2) = z^3
     y - sqrt(-2) = w^3

Now let z = u + v*sqrt(-2), with u and v in Z. Then w = u - 
v*sqrt(-2), and

     y + sqrt(-2) = u^3 + 3*u^2*v*sqrt(-2) - 6*u*v^2 - 2*v^3*sqrt(-2)
     y - sqrt(-2) = u^3 - 3*u^2*v*sqrt(-2) - 6*u*v^2 + 2*v^3*sqrt(-2)

Subtracting the second from the first,

     2*sqrt(-2) = 6*u^2*v*sqrt(-2) - 4*v^3*sqrt(2)
              1 = 3*u^2*v - 2*v^3
              1 = v*(3*u^2-2*v^2)

This implies that v = 1 or -1, and it's easy to see that the only 
solutions are v = 1, u = 1 or -1. Then

     x = w*z = 3

     y = u^3 - 6*u*v^2
       = u*(u^2-6*v^2)
       = -5*u
       = 5 or -5

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
College Modern Algebra
College Number Theory

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