Finding Integer Solutions of x^3 - y^2 = 2Date: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/