No Solution: y^2 = x^3 + 7Date: 03/17/2003 at 10:53:40 From: Weijia Subject: Congruences Show that y^2 = x^3 + 7 has no integer solution. I have received two hints from my professor about how to go about proving this, and the first is to show that if x is even, then the solution does not exist, and the second is that when x is odd, we can write x = 2z +1, and if we plug x = 2z+1 into y^2 = x^3 +7 and then factor out y^2 + 1 = x^3 + 8 = ( ) (), and then show that there exist among the factors a prime p which is congruent to 3 mod 4. For the first part I thought if x is even then that implies y^2 is congruent to 7 mod 8, and that can not happen, but the second part is really beyond me. I have no idea how the hint is supposed to help me. Please enlighten me on this, Dr. Math. Date: 03/18/2003 at 05:37:39 From: Doctor Jacques Subject: Re: Congruences Hi Weijia, The first part is correct - x cannot be even. Assume now that x is odd. In that case, we have x^3 = x mod 4 (by Euler's theorem, as phi(4) = 2). We have: y^2 = x + 7 mod 4 As x is odd, y is even, and y^2 = 0 mod 4, which shows that x = 1 mod 4. We also have, as the hint suggests: y^2 + 1 = x^3 + 8 = (x + 2)(x^2 - 2x + 4) and x + 2 = 3 mod 4. This means that not all the prime factors of x + 2 can be of the form 4k + 1, as we would otherwise have x + 2 = 1 mod 4. We have a prime factor p of x + 2 congruent to 3 mod 4. We can write: y^2 + 1 = 0 mod p and this is impossible by Gauss' Quadratic Reciprocity laws. Does this help? Write back if you'd like to talk about this some more, or if you have any other questions. - Doctor Jacques, The Math Forum http://mathforum.org/dr.math/ |
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