Interesting Diophantine EquationDate: 12/06/2005 at 17:00:57 From: Jenny Subject: Squares and powers of 3 Find all integers x such that x^2 + 3^x is the square of an integer. The numbers can get pretty large pretty quickly, so it's quite hard to check cases of larger values of x. Perhaps it may be possible to factorize the expression, or regard the x^2 and 3^x separately. I'm not really sure where to go on this problem, and help would be greatly appreciated. Date: 12/06/2005 at 22:08:04 From: Doctor Vogler Subject: Re: Squares and powers of 3 Hi Jenny, Thanks for writing to Dr. Math. What you have is a type of Diophantine equation, which means that you want integer solutions to your equation. You want all pairs of integers (x, y) that satisfy x^2 + 3^x = y^2. One idea is to notice that, if x is large, then y must be a lot larger, because 3^x is very big compared to x^2. That means that y^2 must be very close to 3^x. If x is even, then y must be very close to 3^(x/2). So we have Case 1: x is even. Now, x = 0 is a solution, but otherwise, if x = 2k is even, then y^2 = 3^x + x^2 > 3^x = 3^(2k) = (3^k)^2, so y > 3^k. That means that y >= 3^k + 1, y^2 >= 3^(2k) + 2*3^k + 1, and so (2k)^2 = x^2 = y^2 - 3^x = y^2 - 3^(2k) >= 2*3^k + 1 > 2*3^k, or 2k^2 >= 3^k, but when k > 0, this never happens. Case 2: x is odd. Trying to use the above method doesn't work at all when x is odd, because you can find squares that are very close to 3 times another square. But then we have a very different idea that doesn't depend on whether x is even or odd. The idea is to move over the x^2 and factor that side of the equation 3^x = y^2 - x^2 3^x = (y - x)(y + x), Now, if the product on the right is a power of 3, then the only way that can happen is if each factor is a power of 3 (or if each factor is the negative of a power of 3). So that means that y + x = 3^a y - x = 3^b which implies y = (3^a + 3^b)/2 x = (3^a - 3^b)/2 x = a + b So now we just need to solve the equation 2(a + b) = 3^a - 3^b. We can assume that a will be larger than b, so let's define c = a - b and then we have 2(2b + c) = [3^c - 1]*3^b. Now, if we can find that 4b < 3^b and 2c < 3^c - 2 then we will be able to conclude that 4b + 2c < 3^b + 3^c - 2 <= 3^b + (3^b)*(3^c - 2) = 3^b * (3^c - 1) = 2(2b + c), which is, of course, impossible. But it turns out that 4b < 3^b when b >= 2, and 2c < 3^c - 2 when c >= 2. So that means that if we have any solutions at all, they must satisfy b <= 1 c <= 1, but a = c + b <= 2, and x = a + b <= 3, which means that we only have to check the possibilities x = 0, 1, 2, 3. And it turns out that three of the four work. So we have only the three solutions: x = 0, x = 1, x = 3. 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: 12/07/2005 at 14:18:16 From: Doctor Vogler Subject: Re: Squares and powers of 3 Hi Jenny, I just realized that I made a mistake in my last answer to you, so I will correct that here. You'll recall that I said: But it turns out that 4b < 3^b when b >= 2, and 2c < 3^c - 2 when c >= 2. So that means that if we have any solutions at all, they must satisfy b <= 1 c <= 1. This is not quite correct. I should have said: So that means that if we have any solutions at all, they must satisfy b <= 1 or c <= 1. I haven't proven that a solution will satisfy both inequalities. Now, if c <= 0, then this implies that a <= b, and therefore x <= 0. But we already know about the solution x = 0, and x < 0 implies that x^2 + 3^x is not an integer, therefore not the square of an integer. So the only way for c <= 1 is if c = 1. What happens when c = 1? Then 2(2b + c) = [3^c - 1]*3^b becomes 2(2b + 1) = [3 - 1]*3^b = 2*3^b 2b + 1 = 3^b. We already know that 4b < 3^b when b >= 2, so even if c = 1, we still must have b <= 1. So now we know that, apart from x = 0, any solution must satisfy b <= 1. But b < 0 implies that x = (3^a - 3^b)/2 is not an integer. So we conclude that either b = 0 or b = 1. If b = 0, then 2(2b + c) = [3^c - 1]*3^b becomes 2c = 3^c - 1 which happens exactly when c = 1, and then c > 1 makes the right side too big. So we have one solution with b = 0, which is b = 0, c = 1, a = 1, x = (3 - 1)/2 = 1, y = 2. If b = 1, then 2(2b + c) = [3^c - 1]*3^b becomes 4 + 2c = 3^(c+1) - 3 or 7 + 2c = 3^(c+1) which happens exactly when c = 1, again, and then c > 1 makes the right side too big. So we have one solution with b = 1, which is b = 1, c = 1, a = 2, x = (9 - 3)/2 = 3, y = 6. And those are all of the possibilities, so that means that those three solutions x = 0 x = 1 x = 3 are the only ones. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ |
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