One Proof that a Diophantine Equation in Two Variables Has Only Three SolutionsDate: 01/30/2011 at 03:17:22 From: Sandesh Subject: Diophantine Equation with exponents Find all integers x and y for which (2^x) + (3^y) is a perfect square. I found some solutions, but I want a full method to solve it comprehensively. Date: 01/30/2011 at 23:36:45 From: Doctor Vogler Subject: Re: Diophantine Equation with exponents Hi Sandesh, Thanks for writing to Dr. Math. That's not an easy problem for a 14-year-old! To prove that there are only three solutions, you must first familiarize yourself with modular arithmetic ... Mod, Modulus, Modular Arithmetic http://mathforum.org/library/drmath/view/62930.html ... as well as Pythagorean triples Pythagorean triples http://mathforum.org/dr.math/faq/faq.pythag.triples.html Here is how I would prove it: We want to find integer solutions to the equation (2^x) + (3^y) = m^2. First of all, if x or y is negative, then (2^x) + (3^y) won't be an integer. Next, if x = 0, then 3^y = m^2 - 1 = (m - 1)(m + 1). So both m - 1 and m + 1 are powers of 3. Since their difference is 2, the only possibility is m - 1 = 1 and m + 1 = 3. So m = 2 x = 0 y = 1 Next, if y = 0, then 2^x = m^2 - 1 = (m - 1)(m + 1). So both m - 1 and m + 1 are powers of 2. Since their difference is 2, the only possibility is m - 1 = 2 and m + 1 = 4 So m = 3 x = 3 y = 0 Now, if y > 0, then 3^y is divisible by 3, so 2^x = m^2 mod 3. But no squares are 2 mod 3, so that means that x must be even. If x > 0, then x >= 2 since it is even, and therefore 2^x is divisible by 4. So 3^y = m^2 mod 4. But no squares are 3 mod 4, so that means that y must also be even. That means that 2^x + 3^y = m^2 gives you a Pythagorean triple: (2^(x/2), 3^(y/2), m). Obviously, 3^(y/2) is odd, so that means that there are coprime integers r and s such that 2^(x/2) = 2rs, 3^(y/2) = r^2 - s^2, and m = r^2 + s^2. But 2^(x/2) = 2rs means that both r and s must be powers of 2. And since they are coprime, one of them must be 1. But 3^(y/2) > 0, so it must be that s = 1 r = 2^(x/2 - 1). And we conclude that 3^(y/2) = r^2 - s^2 = 2^(x - 2) - 1. Now, 2^(x - 2) is still a square, since x - 2 is even, so we can factor the right side as 3^(y/2) = [2^(x/2 - 1) - 1][2^(x/2 - 1) + 1] And since the product is a power of 3, that means that each term must be a power of 3: 2^(x/2 - 1) - 1 = 3^a 2^(x/2 - 1) + 1 = 3^b But their difference is exactly 2, so the only possibility is that the smaller one is 1 and the larger one is 3: 2^(x/2 - 1) - 1 = 1 2^(x/2 - 1) + 1 = 3 This means that 2^(x/2 - 1) = 2. So x/2 - 1 = 1 x/2 = 2 x = 4 And y = 2. So there are only three solutions, namely (x, y) = (0, 1), (3, 0), and (4, 2). 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: 01/31/2011 at 07:41:55 From: Sandesh Subject: Diophantine Equation with exponents Thank you for a very good rigorous answer. But can you please also tell me the proof of the Pythagorean triples that are of this form? (2rs, r^2 - s^2, r^2 + s^2) Also, give a proof that r and s are coprime. Date: 01/31/2011 at 15:58:43 From: Doctor Vogler Subject: Re: Diophantine Equation with exponents Hi Sandesh, Those can be found at the second link I mentioned earlier: http://mathforum.org/dr.math/faq/faq.pythag.triples.html - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ |
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