Proving a Polynomial is a Perfect Square
Date: 08/17/2007 at 13:47:51 From: Priyanka Subject: proving that a polynomial is a perfect square. Let a, b be odd integers such that a^2 - b^2 + 1 divides b^2 - 1. Prove that a^2 - b^2 + 1 is a perfect square.
Date: 08/21/2007 at 01:51:30 From: Doctor Vogler Subject: Re: proving that a polynomial is a perfect square. Hi Priyanka, Thanks for writing to Dr. Math. That's a really good question. I've been working on it since you posted it on Friday, because it really intrigued me. And I do love Diophantine equations. A Diophantine equation is an equation (usually polynomials in at least two variables) where you want solutions which are integers. In your case, the equation is (a^2 - b^2 + 1)k = b^2 - 1. This equations has lots of solutions, infinitely many solutions. And you want to prove that all of them either have a or b even or a^2 - b^2 + 1 a perfect square. The thing that makes this especially hard to prove is that when a or b is even, a^2 - b^2 + 1 is often not a perfect square! And there are infinitely many solutions where a or b is even. The first thing I noticed is that all of the solutions I found where a and b were both odd had either a = b or had the form a = 4m^3 - m b = 4m^3 - 3m k = m^2 - 1 for some odd number m. Later, I found solutions not of this form, but they were pretty large, so I didn't find them until I understood this problem a lot better. So at first I wondered if all solutions with a and b both odd had this form. In particular, a^2 - b^2 + 1 being a square is equivalent to k + 1 being a square, since adding a^2 - b^2 + 1 to each side of your original equation gives (a^2 - b^2 + 1)(k + 1) = a^2, and a square times a non-square can never equal a square. From the same equation, it would suffice to prove that k + 1 was relatively prime to (had no prime factors in common with) a^2 - b^2 + 1, which seemed to be true, but I was unable to prove this. It seemed better to consider it a kind of Pell Equation. For example, let's consider all integer solutions that have k = 1 (not just the odd ones). These have a^2 - b^2 + 1 = b^2 - 1 or a^2 = 2(b^2 - 1). Since this implies that a^2 is even, that means that a is even, and 2(a/2)^2 = b^2 - 1, or b^2 - 2(a/2)^2 = 1. This equation is familiar to someone who has studied number fields. It says that the norm of b + (a/2)*sqrt(2) is 1. It turns out that all such numbers are powers of 3 + 2*sqrt(2), or the negative of such a power. Since we don't care about the signs of a and b, we can write b + (a/2)*sqrt(2) = (3 + 2*sqrt(2))^n for some n. For example, n = 3 gives (multiplying out the left side) (3 + 2*sqrt(2))^3 = 99 + 70*sqrt(2), and so b = 99 and a/2 = 70, and (a, b, k) = (140, 99, 1) is one solution to your equation. In fact, we can generalize this. You can always transform your equation like so: (a^2 - b^2 + 1)k = b^2 - 1 ka^2 = (k + 1)(b^2 - 1) k(k + 1)(a/(k + 1))^2 = b^2 - 1 b^2 - (k(k + 1)) (a/(k + 1))^2 = 1. Then you'll get a solution b + (a/(k + 1))*sqrt(k(k + 1)) for each power of (2k + 1) + 2*sqrt(k*(k + 1)). For example, for the power n = 1, you get a = 2k + 2 b = 2k + 1. For n = 2, you get a = 8k^2 + 12k + 4 b = 8k^2 + 8k + 1. And if k = m^2 - 1 for some m, then you can transform your equation like so (a^2 - b^2 + 1)(m^2 - 1) = b^2 - 1 (m^2 - 1)a^2 = m^2(b^2 - 1) (m^2 - 1)(a/m)^2 = b^2 - 1 b^2 - (m^2 - 1) (a/m)^2 = 1. Then you'll get a solution b + (a/m)*sqrt(m^2 - 1) for each power of m + sqrt(m^2 - 1). For example, for the power n = 1, you get a = m b = m k = m^2 - 1. For n = 3, you get a = 4m^3 - m b = 4m^3 - 3m k = m^2 - 1. Importantly, if m is odd and n is also odd, then a and b are odd too. It turns out that this will give you all solutions where a and b are odd. So let's get around to proving this: Since we need to work with square roots of k(k + 1), we need to pull out any square factors. For example, if k = 3, then we need to work with sqrt(3) instead of sqrt(12). So let k(k + 1) = rs^2 where r is the square-free part of k(k + 1), and s is the largest square that divides k(k + 1). Since k(k + 1) can't be negative (even if k is), we can assume that r and s are positive. Then we transform the equation into b^2 - r(ak/rs)^2 = 1. So we this means that b + (ak/rs)*sqrt(r) has norm 1. Note that ak/rs is an integer; you see, (ka)^2 = k(k + 1)(b^2 - 1) (ka)^2 = rs^2(b^2 - 1), which means that s^2 divides (ka)^2, and so s divides ka. Also, r divides (ka/s)^2, and since r is square-free, this implies that r divides ka/s, so ka/rs is an integer. By Dirichlet's Unit Theorem, all numbers of the form x + y*sqrt(r) with x and y positive integers, which have norm 1, are powers of a single such number. That is, there are positive integers u and v (depending only on k) such that x + y*sqrt(r) = (u + v*sqrt(r))^n for some integer n. (Actually, Dirichlet's Unit Theorem says more than this, but since r is positive, it implies a fundamental unit in the ring of integers of Q(sqrt(r)), and all numbers with norm 1 will either be powers of this unit or powers of its square, and all numbers with x and y integers will either be powers of this unit (or its square) or powers of the cube. So the number u + v*sqrt(r) is the fundamental unit raised to the 1, 2, 3, or 6.) Now, we have one other important fact. The number (2k + 1) + (2s)*sqrt(r) also has norm 1, since (2k + 1)^2 - r(2s)^2 = (4k^2 + 4k + 1) - 4(k^2 + k) = 1, so it is also a power of u + v*sqrt(r). Now consider what is the largest power of 2 which divides v. Say that it is t, so that v/2^t is an odd integer. Then a little modular arithmetic will show that odd powers of u + v*sqrt(r) will have the form x + y*sqrt(r) where y is divisible by 2^t but not 2^(t+1), but even powers of u + v*sqrt(r) will have y divisible by 2^(t+1). Now, we know that (2k + 1) + (2s)*sqrt(r) is such a power, so the question is whether 2^(t+1) divides 2s. If it does, then that means that (2k + 1) + (2s)*sqrt(r) is an even power of u + v*sqrt(r), so it is the square of some other such power, say (2k + 1) + (2s)*sqrt(r) = (x + y*sqrt(r))^2 = (x^2 + ry^2) + (2xy)*sqrt(r) where x^2 - ry^2 = 1, but then 2k + 1 = x^2 + ry^2 = 2x^2 - 1, and so k + 1 = x^2 is a perfect square, as we needed to prove. The alternative is 2^(t+1) does not divide 2s, but only 2^t. But then that means that b + (ak/rs)*sqrt(r), which is also a power of u + v*sqrt(r), must have 2^t divides (ak/rs). But ak/rs = as/(k + 1), so 2^t divides as/(k + 1). But if a and b are both odd, then k is even, and so k + 1 is odd. Therefore, 2^t divides as. But since 2^(t+1) does not divide 2s, that means that 2^t does not divide s, and therefore a must be even. 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/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2015 The Math Forum