No Integer SolutionDate: 10/21/2001 at 03:51:21 From: Tomas Prochazka Subject: z^2-2y^2=51 I don't know how to prove that the following equation has no solution in Z (integers): z^2 - 2y^2 = 51 Another equation was z^2 - 4y^2 = 51 (z-2y)(z+2y) = 51 and both () must be integers and 51 = 1*3*17 => z-2y = 3 z+2y = 17 ... This is easy, but what to do with the first one? I think that (z-2^(1/2)y)(z+2^(1/2)y) cannot be used because brackets aren't integers, so factoring 51 won't help. Thanks, Tom Date: 10/21/2001 at 05:51:08 From: Doctor Mitteldorf Subject: Re: z^2-2y^2=51 Dear Tom, There is, actually at least one pair of squares with a difference of 51, and it's not hard to find. If you write a list of squares, 1 4 9 16 25 36 ... you'll notice that the difference between successive squares is a complete series of odd integers: 3 5 7 9 11 ... How far will you have to go until you have two successive squares have a difference of 51? There's one more pair that you'll find pretty easily by trial-and- error: 10 and 7. Could you have predicted this from an equation? How about trying (x+3)^2 - x^2 = 51? Are there any others? What happens if you write (x+5)^2 - x^2 = 51? - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/ Date: 10/21/2001 at 05:59:17 From: Tomas Prochazka Subject: Re: z^2-2y^2=51 Differences between squares are odd numbers; when you compare x^2 and (x+1)^2 the difference is 2x+1, which is an odd number; if it is x^2 - y^2 = 51 then you're right, and it is very simple. But I have x^2 - 2*y^2 = 51 and y is 2*, so I think it is more difficult. I tried to write a program which tries numbers from -1000 to 1000 and it didn't find any solution. Tomas Prochazka Date: 10/21/2001 at 07:18:40 From: Doctor Mitteldorf Subject: Re: z^2-2y^2=51 Tom - Just now in the shower, I was thinking about an interesting generalization of the question you raised. Here's a challenge for you: We just saw that 51 can be expressed as the difference of two squares in two different ways. What's the smallest number that can be expressed as the difference of two squares in three different ways? The kind of algebraic thinking we've been doing should put us on a fast track to an answer. Will you write to me with yours? - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/ Date: 10/21/2001 at 07:43:26 From: Tomas Prochazka Subject: Re: z^2-2y^2=51 Dr. Mitteldorf, I tried to find a number that is the difference of 2 roots whose smallest difference (x-n)^2 - x^2 where n is smallest in three ways and I think 45 can be the result: (x+1)^2-x^2 = 2x+1 (x+3)^2-x^2 = 6x+9 (x+5)^2-x^2 = 10x+25 (x+n)^2-x^2 = 2nx+n^2 so that number must be odd (2x+1) and after dividing it by 6 the remainder must be 3; the first two numbers that fulfill this requirement are 15 and 45. But for 15 n = -1 and n = 1, which gains the same result twice (4^2 - 1^2 = 15), so 45 won: 23^2-22^2 = 45 9^2-6^2 = 45 7^2-2^2 = 45 but I still don't know what to do with: x^2 - 2*y^2 = 51 Tom Date: 10/23/2001 at 16:40:18 From: Doctor Mitteldorf Subject: Re: z^2-2y^2=51 Dear Tomas, I'm sorry - I misread your question. Thanks for writing back. Your analysis of the challenges that I posed was very smart, and I think you are right, that 45 is the smallest number that can be expressed as the difference of two squares in two different ways. The case of x^2 - 2y^2 = 51 is more difficult. I think I have a proof from mod 3 arithmetic. What is x mod 3? Suppose x mod 3 = 0. Then y mod 3 must also be 0, so that the difference x^2 - 2y^2 is also divisible by 3. But this cannot be, because x^2 and y^2 would both be divisible by 9; hence x^2 - 2y^2 is divisible by 9, but 51 is not. Suppose x mod 3 = 1. Then x^2 mod 3 = 1, and 2y^2 mod 3 must = 1, so y^2 mod 3 must = 2. But 2 is not a square mod 3. There is no way to satisfy y^2 mod 3 = 2. Suppose x mod 3 = 2. Then x^2 mod 3 = 1, and the same argument applies. There are no other cases. Please check this proof for me - I think it is right, but I am not sure. - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/ Date: 10/24/2001 at 17:55:34 From: Tomas Prochazka Subject: Re: z^2-2y^2=51 I think you're right. I tried to write a program that checked [X,Y] from <-1000;1000>x<-1000;1000> and no two numbers were solution. Tom |
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