Interesting Integer Problem with Diophantine EquationsDate: 06/21/2005 at 11:19:09 From: Shamik Subject: Number Theory: Finding out two positive numbers..... Two positive integers are such that the difference of their squares is a cube and the difference of their cubes is a square. Find the smallest possible numbers that satisfy the same and a way to find all such pairs of positive integers (a, b) that would work. Let the two positive numbers be a and b, with a > b. Now, a^2 - b^2 = m^3 ==> (a+b)*(a-b) = m^3 It implies that (a-b) = m and (a-b)^2 = m^2. a = (m^2 + m)/2 and b = (m^2 - m)/2. Now, there would be an infinite series satisfying this condition that follows: m a b ------------------ 1 1 0 2 3 1 3 6 3 4 10 6 5 15 10 6 21 15 7 28 21 8 36 28 9 45 36 10 55 45 and, so on. But, a^3 - b^3 = n^2 => n^2 = (a-b)*(a^2 + ab + b^2) = (m^2/4)*(3*m^3 + m) So, {3*(m^3) + m} should be a square as (m^2/4) is of that form. I am completely stuck at this point. Please, help me out. I could find out the following by just trial and error. The lowest possible solution is (a,b) = (10,6) 10^2 - 6^2 = 64 = 4^3 10^3 - 6^3 = 784 = 28^2 But, I am nowhere near to finding a generic solution. Eagerly waiting for your reply. Thanks and regards, Shamik Date: 06/21/2005 at 18:25:04 From: Doctor Pete Subject: Re: Number Theory: Finding out two positive numbers..... Hi Shamik, If we have a^2 - b^2 = m^3, a > b > 0, then (a-b)(a+b) = m^3, but I don't see how this implies a-b = m. Indeed, for m = 3, we find a = 14, b = 13, since 14^2 - 13^2 = (14-13)(14+13) = 27 = 3^3. In fact for any positive integer k, a^2 - b^2 = (2k+1)^3 admits the solution a = 4k^3 + 6k^2 + 3k + 1, b = 4k^3 + 6k^2 + 3k. This includes a class of solutions to the first condition that is not among those you mentioned. - Doctor Pete, The Math Forum http://mathforum.org/dr.math/ Date: 06/22/2005 at 00:40:31 From: Shamik Subject: Number Theory: Finding out two positive numbers..... Hi Doctor Pete, Thanks for your reply. It was a mistake that I overlooked the other class of solutions. As I was not able to proceed on pure mathematical grounds, I used computer programming to get hold of a few answers. Case (1): When a - b = 1, we have the following answers: a b ------------------ ------------------ 1 0 296247805703169927 296247805703169926 354103272035509784 354103272035509783 Case (2): When (a - b) is NOT = 1, we have the following answers: a b ------------- ------------ 1 0 10 6 50005000 49995000 800020000 799980000 4050045000 4049955000 12800080000 12799920000 31250125000 31249875000 64800180000 64799820000 120050245000 120049755000 204800320000 204799680000 328050405000 328049595000 There are lots more of them. Is there any mathematical way to arrive at these answers? Or is computer programming the only way we can think of? Please, give me some directions. Thanks and regards, Shamik Date: 06/23/2005 at 05:50:36 From: Doctor Pete Subject: Re: Number Theory: Finding out two positive numbers..... Hi Shamik, I don't have an answer for you, but I have conducted my own computer search and found the following solutions, which are minimal in the sense that there are no other pairs (a,b) for which a^2 - b^2 = m^3 for m <= 1000: (10, 6), (640, 384), (7290, 4374), (8954, 5687), (70434, 64350). Notice that all but the first solution are not listed among any of the ones you have found so far. - Doctor Pete, The Math Forum http://mathforum.org/dr.math/ Date: 06/30/2005 at 19:57:40 From: Doctor Vogler Subject: Re: Number Theory: Finding out two positive numbers..... Hi Shamik, I thought I would say a few things about your question, such as how I would attack it. You have two simultaneous Diophantine equations x^2 - y^2 = z^3 x^3 - y^3 = w^2. Whenever you have a Diophantine problem like this, there are several questions you try to answer. The first is: Is there a solution? If you can't find one, then you try to prove that there is no solution. But you already found the solution (10, 6, 4, 28), so you know the answer to this question is yes. In that case, you go on to the second question: Are there only finitely many solutions, or infinitely many? If you think there are only finitely many, then you try to find all of them, and then you try to prove that you have all of them. If you think there are infinitely many, then you try to prove this, usually by either coming up with a formula to get from one solution to a bigger solution, or by coming up with a formula in terms of some variable n that gives a different solution for every integer n. Neither of those is possible for every problem. But it is for yours. Finally, you try to show that your formula gives you all integer solutions. Again, this is not always possible. For your problem, it is easy to prove that any of the particular formulas I come up with does NOT cover all of the solutions, simply by finding one not in the list, but it is much harder to prove that there is no formula that DOES cover all of the solutions. As for answering the question of whether there are finitely many solutions or infinitely many, it is helpful to have a computer try to find some solutions. Doctor Pete already listed off some (10, 6), (640, 384), (7290, 4374), (8954, 5687), (70434, 64350), and there are more; they just keep getting bigger. So we might be led to suspect that there are infinitely many solutions. (We will see later that this is true.) Now let's start out by seeing what we can say about ALL solutions. At some point, we'll have to take a leap and give up covering all of them, but it's good to see how far you can get trying not to miss any (especially since you might miss all of the solutions if you make too many assumptions too soon). (Of course, you have the infinitely many solutions (10n^6, 6n^6), but that seems like cheating; in some sense that's only one solution.) So let's see if we can find all integer solutions to x^2 - y^2 = z^3. The left side factors as (x - y)(x + y) = z^3. Now let's consider what x - y might be. Let m = x - y. (Notice that x^3 - y^3 = w^2 >= 0 implies that m >= 0. And m = 0 is always a solution.) Then m(2x - m) = z^3. So clearly m divides z^3. What does that say about z? Well, if m=1, it doesn't mean anything. But if m=2 or m=3, or if m is any squarefree integer, then the only way that m can divide z^3 is if m divides z. But that's not true for m=4. In general, if we let a be the largest number whose cube divides m, then m/a^3 will be a product of primes raised to the first and second powers. If we group those into two numbers c and b^2, then we have m = a^3 * b^2 * c where b and c are squarefree, relatively prime positive integers. And then m divides z^3 exactly when abc divides z. So let k = z/abc, and then 2x - m = z^3/m 2x - a^3*b^2*c = k^3*b*c^2 2x = k^3*b*c^2 + a^3*b^2*c x = bc(k^3*c + a^3*b)/2. And this will be an integer exactly when b is even, or c is even, or a and k have the same parity (both even or both odd). And then x = bc(k^3*c + a^3*b)/2, y = bc(k^3*c - a^3*b)/2, z = kabc. Well! We're making progress! Now we have a nice expression for all solutions to the first equation. We let b and c run over all pairs of relatively prime squarefree positive integers, and let a and k run over all integers, and that will give us all solutions to the first equation, x^2 - y^2 = z^3. Now we need to decide which of those solutions also provide solutions to x^3 - y^3 = w^2. So if we substitute our expressions for x and y into this equation, and simplify, we get a^3*b^4*c^3*(a^6*b^2 + 3c^2*k^6) = 4w^2. That doesn't look pretty. But right away we can say that a^3*b^4*c^3 divides (2w)^2, which tells us that a*b^2*c divides 2w. In fact, since c is squarefree, that means that a*b^2*c^2 divides 2w. But then this implies that (a*b^2*c^2)^2 divides the right side of the equation, and we can already see that it divides the second term on the left side, namely a^3*b^4*c^3*(3c^2*k^6), which means that it also has to divide the first term a^3*b^4*c^3*(a^6*b^2), which means that c divides a^7*b^2. But b and c are relatively prime, so that means that c divides a^7. And c is squarefree, which means that this can only happen if c divides a. So if we let d = a/c, then we have c^3*d^3*b^4*c^3*(c^6*d^6*b^2 + 3c^2*k^6) = (2w)^2, and we find that b^4*c^8*d^3 divides (2w)^2, so that r = 2w/(b^2*c^4*d) is an integer. Then our equation becomes d(c^4*d^6*b^2 + 3k^6) = r^2. Now we have the same kind of issue that we had before with m dividing z^3; if d is squarefree, then d dividing r^2 implies that d divides r. But if d is not squarefree, then it does not. As before, we can write d = s^2*t where t is squarefree, by dividing by the largest possible square, and then d dividing r^2 implies that st divides r. So now we can write r = stu, and this turns our equation into t*u^2 = b^2*c^4*t^6*s^12 + 3k^6. The thing to do at this point is to use the fact that t must divide 3k^6. But then you have to split into two cases, one for t divisible by 3, and one for t not divisible by 3. (In the second case, remember that t is squarefree, so this implies that t divides k. Very nice.) But those two equations don't help us very much anyway, so I will stop here and change gears. If you want a program that will find all solutions, then I might recommend starting with this equation (or the two cases, where t divides k and where t/3 divides k). Then substitute back to get x and y. This will allow you to get many more solutions with much smaller numbers. Also, it might be helpful to change b^2 * c^4 * t^6 * s^12 back to a single number q^2, but the only trouble with this is that you have to find the prime divisors of q whose exponents (in the prime factorization of q) are 3, 4, or 5 mod 6, since the product of these primes is the number t, which appears elsewhere in the equation. Of course, you only have to check primes up to m^(1/3), so this isn't all that much work. But now let's take a leap. We suspect that there are infinitely many solutions, so maybe we won't lose too many if we make an assumption like t = 1. When we do this, then we can factor the equation as (u + b*c^2*s^6)(u - b*c^2*s^6) = 3k^6. Now all we need to do is find some factor f of 3k^6, and set g = 3k^6/f, and then solve u + b*c^2*s^6 = f u - b*c^2*s^6 = g for u = (f + g)/2 b*c^2*s^6 = (f - g)/2. In this case, we get r = su = su d = s^2 a = cd = cs^2 w = a*b^2*c^3*r/2 = b^2*c^4*s^3*u/2 x = (1/2)*b*c^2*(k^3 + b*c^2*s^6) y = (1/2)*b*c^2*(k^3 - b*c^2*s^6) z = b*c^2*s^2*m. But then how will we know the factorization of f - g? We need to be able to distinguish s^6 from b*c^2, at least. And we need that all of the exponents of prime factors of f - g are 0, 1, or 2 mod 6 (and not 3, 4, or 5), since otherwise we wouldn't have t = 1. But if we just picked some random f and some random g, then it probably won't have very many cubes dividing it. So let's just suppose that f - g is cubefree. Then that gives us s = 1, and b*c^2 = (f - g)/2. We substitute that and get x = (1/2)*(f - g)/2*(k^3 + (f - g)/2) y = (1/2)*(f - g)/2*(k^3 - (f - g)/2) z = k*(f - g)/2 w = (f - g)^2(f + g)/16 [(1/2)*(a - b)/2*(c^3 + (a - b)/2),(1/2)*(a - b)/2*(c^3 - (a - b)/2), c*(a - b)/2,(a - b)^2*(a + b)/16] Of course, we also need f and g to have the same parity (both even or both odd) in order that these numbers be integers. But now let's see if we can find some factors f and g of three times sixth powers 3k^6, and see what we get. If we let f = 3n^2, g = n^4, k = n, then we get polynomials in n for x, y, z, and w, namely x = 1/8*n^8 - 1/4*n^7 - 3/4*n^6 + 3/4*n^5 + 9/8*n^4, y = -1/8*n^8 - 1/4*n^7 + 3/4*n^6 + 3/4*n^5 - 9/8*n^4, z = -1/2*n^5 + 3/2*n^3, w = 1/16*n^12 - 3/16*n^10 - 9/16*n^8 + 27/16*n^6. And these polynomials satisfy your two equations, which means that, for any number n, they will still satisfy the equations. The coefficients are not all integers, but if you substitute n = 2m, then you have all integer coefficients (so all even numbers give integer results) and if you substitute n = 2m+1 then you have all integer coefficients (so all odd numbers give integer results). We can do the same thing by substituting other combinations like f = 3n, g = n^5, k = n (which gives integers results when n is odd or divisible by 4), and f = 3n^6, g = 1, k = n (which requires n to be odd). But you might complain that all of these give polynomials in y with negative leading terms, meaning that most y values will be negative. If you want positive values for y, then you can try something like f = 3n^6, g = m^6. You'll need n and m to be fairly close to one another (but of the same parity), so try something like m = n + 2, or m = n + 4, or something like m = 9h, n = 7h + 2, or something similar. For example, m = n + 2 gives you x = n^12 - 6*n^11 - 39*n^10 - 62*n^9 + 306*n^8 + 2016*n^7 + 6000*n^6 + 11520*n^5 + 15264*n^4 + 13952*n^3 + 8448*n^2 + 3072*n + 512 y = 6*n^11 - 15*n^10 - 262*n^9 - 1314*n^8 - 4032*n^7 - 8688*n^6 - 13824*n^5 - 16416*n^4 - 14208*n^3 - 8448*n^2 - 3072*n - 512 z = n^8 - 4*n^7 - 42*n^6 - 140*n^5 - 280*n^4 - 336*n^3 - 224*n^2 - 64*n w = n^18 - 9*n^17 - 45*n^16 - 12*n^15 + 1440*n^14 + 12276*n^13 + 63132*n^12 + 235584*n^11 + 676368*n^10 + 1534720*n^9 + 2787840*n^8 + 4068864*n^7 + 4751616*n^6 + 4386816*n^5 + 3133440*n^4 + 1671168*n^3 + 626688*n^2 + 147456*n + 16384 You might also find that most of your solution polynomials factor nicely too. Take any one of these, and then every integer n will give you a different solution. Well, there might be a few repeats, but you'll still get infinitely many solutions. 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: 07/04/2005 at 04:23:40 From: Shamik Subject: Thank you (Number Theory: Finding out two positive numbers.....) Thank you very much. The process you have shown is wonderful. I was on a leave away from computers so it took me some time to thank you. I thank you very much. Regards, Shamik Date: 07/04/2005 at 12:23:19 From: Doctor Vogler Subject: Re: Thank you (Number Theory: Finding out two positive numbers.....) Hi Shamik, Here's another idea. Recall that we started with x^2 - y^2 = z^3, and x^3 - y^3 = w^2. And then we defined the variables m = x - y, m = a^3 * b^2 * c, with b and c relatively prime squarefree and positive, k = z/abc, so that x = bc(k^3*c + a^3*b)/2, y = bc(k^3*c - a^3*b)/2, z = kabc. Then we defined the integer variables d = a/c, r = 2w/(b^2*c^4*d), d = s^2*t with t squarefree, and u = r/st. This finally left us with the equation t*u^2 = b^2*c^4*t^6*s^12 + 3k^6, and I suggested: The thing to do at this point is to use the fact that t must divide 3k^6. But then you have to split into two cases, one for t divisible by 3, and one for t not divisible by 3. The idea here is that if t is not divisible by 3, then we see that t divides 3k^6, and therefore t divides k^6. But since t is squarefree, that means that t divides k. Therefore, m = k/t is an integer, and then t^6 divides the right side of the equation, which means that t^5 divides u^2, which means (since, again, t is squarefree) t^3 divides u, so that v = u/t^3 is an integer. Then we have tv^2 = b^2*c^4*s^12 + 3m^6. The nice thing here is that no variable appears in more than one term. That means that we can find all of the solutions where t is not divisible by 3 in the following way. Let b and c run over all relatively prime squarefree positive numbers, and let s and m run over all integers. Compute the sum b^2*c^4*s^12 + 3m^6 and then factor it into prime factors. Let v be the largest square divisor, and let t be the remaining squarefree part. Then define u = v*t^3, k = m*t, r = u*s*t, d = t*s^2, a = d*c, w = (r*b^2*c^4*d)/2, x = bc(k^3*c + a^3*b)/2, y = bc(k^3*c - a^3*b)/2, z = kabc. And you can throw out all of those solutions where the division by 2 doesn't give you an integer. This will give you all solutions where t is not divisible by 3 (and perhaps some where it is, but you can throw those out so as not to duplicate them in the next part). Now what happens when t is divisible by 3? Then we have t = 3*t', and t' divides k^6, so t' divides k, and then t'^5 divides u^2, and now we have m = k/t' v = u/t'^3 t'*v^2 = 3^5*b^2*c^4*s^12 + m^6. Now we let b and c run over all relatively prime squarefree positive numbers, and let s and m run over all integers. Compute the sum 3^5*b^2*c^4*s^12 + m^6 and then factor it into prime factors. Let v be the largest square divisor, and let t' be the remaining squarefree part. If t' is not divisible by 3, then define u = v*t'^3, k = m*t', t = 3*t', r = u*s*t, d = t*s^2, a = d*c, w = (r*b^2*c^4*d)/2, x = bc(k^3*c + a^3*b)/2, y = bc(k^3*c - a^3*b)/2, z = kabc. And this will give us all of the other solutions. In each case, the only hard part is the factoring. But at least you only have to look for primes up to the cube root of vt^2, since if p > N^(1/3) and p^2 divides N, then you would have already found and factored out M = N/p^2 < N^(1/3). - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ Date: 07/05/2005 at 00:55:54 From: Shamik Subject: Thank you (Number Theory: Finding out two positive numbers.....) Doctor Vogler, Thank you very much. I am in a position to understand your excellent approach but then this is the first time I am learning the process. I couldn't have contemplated going in the direction that you have described. Thank you once again. Regards, Shamik |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/