Arithmetic Progression ProofDate: 07/19/2006 at 04:59:21 From: sheryl Subject: can you show me the proof please.. Can you please show me the proof that the product of four positive integers in an arithmetic progression cannot be the square of an integer? Thanks! Date: 07/19/2006 at 14:46:20 From: Doctor Vogler Subject: Re: can you show me the proof please.. Hi Sheryl, Thanks for writing to Dr. Math. I don't know of an easy way to prove this, so I will sketch the proof that I can manage, and you are welcome to ask for more details of any part that is confusing. But I will deliberately leave out many details so that you can work through them yourself. Let's suppose that you have four positive integers in an arithmetic progression, x, x+y, x+2y, x+3y, and their product is a square. You'd like to prove that this is impossible. First of all, explain why you can assume that x and y are relatively prime (i.e. gcd(x,y) = 1). Next, we'd like to say that if x and y are relatively prime and x(x+y)(x+2y)(x+3y) is a square, then each of the factors x, x+y, x+2y, x+3y, are squares. This is not true. However, you can prove that gcd(x, x+y) = 1 gcd(x+y, x+2y) = 1 gcd(x+2y, x+3y) = 1 gcd(x, x+2y) = 1 or 2 gcd(x+y, x+3y) = 1 or 2 gcd(x, x+3y) = 1 or 3. Furthermore, gcd(x, x+2y) and gcd(x+y, x+3y) can't both be 2. Now explain why this implies that there are integers a, b, c, and d (which are pairwise relatively prime) such that one of the following must be true: (1) x = a^2, x+y = b^2, x+2y = c^2, x+3y = d^2 (2) x = a^2, x+y = 2b^2, x+2y = c^2, x+3y = 2d^2 (3) x = 2a^2, x+y = b^2, x+2y = 2c^2, x+3y = d^2 (4) x = 3a^2, x+y = b^2, x+2y = c^2, x+3y = 3d^2 (5) x = 3a^2, x+y = 2b^2, x+2y = c^2, x+3y = 6d^2 (6) x = 6a^2, x+y = b^2, x+2y = 2c^2, x+3y = 3d^2 You'll see this same technique also used in a different problem here: Consecutive Integer Proof http://mathforum.org/library/drmath/view/65589.html Now, three of these cases are easy to eliminate using modular arithmetic. For example, in (3), you have 3b^2 - 4a^2 = d^2. Considering this equation mod 3, you'll find that the only solutions have a and d both divisible by 3. But then 3b^2 is divisible by 9, so b is also divisible by 3. This means that both x and y are divisible by 3, contrary to our hypothesis. You can use a similar argument to rule out (2) and (4). Cases (5) and (6) are actually very similar, by reversing the order of {a, b, c, d}, so I'll just treat them as one. That means that we still need to worry about cases (1) and (5). Well, we can't eliminate either of them by modular arithmetic, because they both have solutions. The quadruple (1, 1, 1, 1) solves (1), and (1, 1, 1, 0) solves (5). Well, in each case, you have an intersection of two quadratic surfaces with a rational point, and this is equivalent to an elliptic curve (or some would say that it *is* an elliptic curve). For example, the equation a^2 = b^2 - d^2 implies that a^2 = (b - d)(b + d) and since b and d are relatively prime, that means that either b + d = r^2, b - d = s^2, a = rs or b + d = 2r^2, b - d = 2s^2, a = 2rs. The other equation defining (5) is c^2 = b^2 + 3d^2. Substituting our formulas, we get either (c/2)^2 = r^4 - r^2*s^2 + s^4, which I suspect you'll find very familiar, or c^2 = 2(r^4 - r^2*s^2 + s^4), which is easy to see has no solutions (recall that r and s can't both be even, and conclude that c^2 is divisible by 2 but not 4). The familiar equation has (as we determined previously) eight families of solutions, all of which give either d = 0 or a = 0, and these do not give positive numbers in your arithmetic sequence. You can do a similar analysis on case (1), and I think you'll find that a^2 = b^2 = c^2 = d^2, which means that the only solutions to your problem are from the constant arithmetic sequence where y = 0. If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ Date: 07/20/2006 at 05:57:10 From: sheryl Subject: can you show me the proof please.. Thank you Dr. Math for a very impressive answer to my question. It helped me a lot! |
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