From: Pierre Bornsztein Subject: Re: Two number theory problems Date: Wed, 12 Jan 2000 20:54:09 +0100 Newsgroups: sci.math Summary: No integer solutions to x^4 -2y^4 = 1 > 2. Show that the Diophantine equation x^4 -2y^4 = 1 > has no nontrivial solutions. (No ideals, no complex > numbers, etc. Probably an infinite descent. An argument > based on a specific modulus won't work, since there is > a trivial solution (1,0).) Hi, Suppose that x,y are integers such that x^4 -2y^4 = 1 (1) From the symmetry of the problem, we may suppose that x,y are positive. Moreover, it's obvious that x is odd. Let x = 2k + 1, where k is a nonnegative integer. From (1), we then have : 2y^4 = x^4 - 1 = (x-1)(x+1)(x^2 + 1) = 8k(k+1)(2k^2 +2k + 1) thus y^4 = 4k(k+1)(2k^2 + 2k + 1) (2) First case : if k is odd. 1) Then k , 4(k+1) , and 2k^2 + 2k +1 are pairwise coprimes. Indeed : - suppose p is a prime which divides k and 4(k+1) then p is odd and divides 4(k+1) - 4k = 4 a contradiction. - suppose p is a prime which divides 4(k+1) and 2k^2 + 2k +1 then p is odd and divides 2(2k^2 + 2k +1) - 4k(k+1) = 2 A contradiction. - suppose p is a prime which divides k and 2k^2 + 2k +1 then p divides 2k^2 + 2k +1 - 2(k+1)k = 1 A contradiction. 2) From (2), it follows that k , 4(k+1) , and 2k^2 + 2k +1 are all perfect fourth powers. Thus 4(k+1) = a^4 and k = b^4 where a,b are nonnegative integers. Then 4b^4 + 4 = a^4 that is (a^2 - 2b^2)(a^2 + 2b^2) = 4 Then a^2 + 2b^2 and a^2 - 2b^2 are two positive divisors of 4, with a^2 + 2b^2 >= a^2 - 2b^2. Thus # a^2 +2b^2 = a^2 -2b^2 = 2 It follows that b = 0 then k = 0 which is impossible since k is odd. # a^2 +2b^2 = 4 and a^2 - 2b^2 = 1 then 2a^2 = 5 which is impossible. then there is no solution in this case. Second case : if k is even. As above, we prove thatk , k+1 , 2k^2 + 2k +1 are pairwise coprimes. It follows that they are all perfect fourth powers. Thus k+1 = a^4 and 4k = b^4. Then 4a^4 = b^4 + 4 that is (2a^2 - b^2)(2a^2 + b^2) = 4 With 2a^2 - b^2 =< 2a^2 + b^2 then # 2a^2 - b^2 = 2a^2 + b^2 = 2 We deduce that b = 0 and k = 0. thus x = 1. Conversely if x = 1 we have y = 0. # 2a^2 - b^2 = 1 and 2a^2 + b^2 = 4 then 4a^2 = 5 which is impossible. Conclusion : the equation x^4 - 2y^4 =1 has only two solutions : (1,0) and (-1;0). Pierre. ============================================================================== From: Pierre Bornsztein Subject: Re: Two number theory problems Date: Wed, 12 Jan 2000 20:56:03 +0100 Newsgroups: sci.math Pierre Bornsztein wrote: > > > 2. Show that the Diophantine equation x^4 -2y^4 = 1 > > has no nontrivial solutions. (No ideals, no complex > > numbers, etc. Probably an infinite descent. An argument > > based on a specific modulus won't work, since there is > > a trivial solution (1,0).) > > Hi, > > Suppose that x,y are integers such that > x^4 -2y^4 = 1 (1) > > From the symmetry of the problem, we may suppose that x,y are positive. Oops. Read x,y are nonnegative... Pierre. ============================================================================== From: Ortwin Gasper Subject: Re: Two number theory problems Date: Wed, 12 Jan 2000 23:55:17 +0100 Newsgroups: sci.math Bart Goddard wrote: > > Hi folks, > > Here are two problems from Ken Rosen's _Elementary > Number Theory_ which I am unable to solve using the > tools provided. I'd appreciate any help. > 2. Show that the Diophantine equation x^4 -2y^4 = 1 > has no nontrivial solutions. (No ideals, no complex > numbers, etc. Probably an infinite descent. An argument > based on a specific modulus won't work, since there is > a trivial solution (1,0).) > Hi Bart, IŽll give you a hint: Let x[n+1] = 2*x[n]+x[n-1], x[0]=1,x[1]=1 and y[n+1] = 2*y[n]+y[n-1], y[0]=0,y[1]=1. Prove, that x[n]^2 - 2*y[n]^2 = (-1)^n for all n in No. (Induction) Show, that these sequences contain _every_ solution of |x^2 - 2*y^2| = 1 in non negative integers. (Look at linear combinations of solutions which are solutions again.Construct a solution smaller than the smallest solution not in the sequence.) Show, that x[2*n] = (2*y[n])^2 + (-1)^n (Look at the explicit formulae of the sequences, wich have the form a*(1+sqrt(2))^n + b*(1-sqrt(2))^n). => Only x[0] is a square of a natural number. => 1^4 - 2*0^4 = 1 is the only (and trivial solution) Regards Ortwin ============================================================================== From: David G Radcliffe Subject: Re: Two number theory problems Date: 13 Jan 2000 01:05:09 GMT Newsgroups: sci.math Bart Goddard wrote: : 2. Show that the Diophantine equation x^4 -2y^4 = 1 : has no nontrivial solutions. (No ideals, no complex : numbers, etc. Probably an infinite descent. An argument : based on a specific modulus won't work, since there is : a trivial solution (1,0).) Clearly x must be odd. (y^2/2)^2 = (x^2-1)/4 * (x^2+1)/2, and the factors on the right are relatively prime integers, so x^2-1 is a square, hence x=+-1. -- David Radcliffe radcliff@alpha2.csd.uwm.edu