Finding Integer Solutions to Four-Variable Equation
Date: 06/10/2007 at 02:54:08 From: Shamik Subject: Number Theory: (a^2 + b^2 + c^2 + d^2) = a*b*c*d a, b, c and d are four natural numbers such that a^2 + b^2 + c^2 + d^2 = a*b*c*d. Find all values of a, b, c and d where all of them are different integers. Obviously, a = b = c = d = 2 is a solution as 2^2 + 2^2 + 2^2 + 2^2 = 16 = 2*2*2*2. I tried to use (A^2 + B^2)*(C^2 + D^2) = (AC - BD)^2 + (AD + BC)^2 = (AD - BC)^2 + (AC + BD)^2 I also had a look at http://www.alpertron.com.ar/4SQUARES.HTM which says "Every positive integer is a sum of four integer squares." Say, N =(a^2 + b^2 + c^2 + d^2)*(A^2 + B^2 + C^2 + D^2) = (aA + bB + cC + dD)^2 + (aB - bA + cD - dC)^2 + (aC - bD - cA + dB)^2 + (aD - dA + bC - cB)^2 Now, if I expand this (aA + bB + cC + dD)*(aB - bA + cD - dC)*(aC - bD - cA + dB)*(aD - dA + bC - cB), will it be equal to N = (a^2 + b^2 + c^2 + d^2)*(A^2 + B^2 + C^2 + D^2) in some special cases? I am asking this as I am not in a position to continue further. Please, set me on the right track with further help if I'm heading towards the wrong direction. I was exploring with the numbers: (i) (1,2,3,4) S = 1^2 + 2^2 + 3^2 + 4^2 = 30 P = 1*2*3*4 = 24 (ii) (1,2,4,5) S = 1^2 + 2^2 + 4^2 + 5^2 = 46 P = 1*2*4*5 = 40 (iii) (1,2,11,12) S = 1^2 + 2^2 + 11^2 + 12^2 = 270 P = 1*2*11*12 = 264 (iv) (1,2,29,30) S = 1^2 + 2^2 + 29^2 + 30^2 = 1746 P = 1*2*29*30 = 1740 (v) (1,2,100,101) S = 1^2 + 2^2 + 100^2 + 101^2 = 20206 P = 1*2*100*101 = 20200 This was the closest I could get where S - P = 6 but I could never get S = P. I understand what I posted above is just a kind of experimental data. It doesn't prove or disprove anything. Please, help me. Thanks and regards, Shamik
Date: 06/11/2007 at 16:43:35 From: Doctor Vogler Subject: Re: Number Theory: (a^2 + b^2 + c^2 + d^2) = a*b*c*d Hi Shamik, Thanks for writing to Dr. Math. The trick to solving this problem is noticing that you can get from one solution to another by replacing d by a*b*c-d. In fact, there are three ways to get new integer solutions from a known integer solution: (1) Change the signs of two of the four variables (2) Permute (or reorder) the four variables (3) Replace d by a*b*c - d. Of course, if any variable is zero, then they must all be zero, and the all-zero solution is unchanged by all three of the above actions. Action (1) always commutes with actions (2) and (3), and it's easy to prove that you must have an even number of positive variables, so you can really just about ignore (1). But (2) and (3) are very very important. For example, we have the solution (2, 2, 2, 2) that you mentioned. Using (3), we get the solution (2, 2, 2, 6). Using (2) and then (3), we get the new solution (2, 2, 6, 22). Using (2) and then (3), we can get either of the solutions (2, 2, 22, 82) or (2, 6, 22, 262). Here we have a solution with a < b < c < d, so this answers your first question that it is possible. Next, using (2) and (3) on those solutions, we can get five more solutions: (2, 2, 82, 306) (2, 22, 82, 3606) (2, 6, 262, 3122) (2, 22, 262, 11522) (6, 22, 262, 34583). Then we can use (2) and (3) on those five solutions to get 14 more, and so on to infinitely many integer solutions (one at each level will have a=b=2, but all of the others will have four different values). Next, you should try to prove that these are all integer solutions. To do this, first you should prove that from any integer solution (except for the all-zero solution), you can use (1), (2), and (3) a finite number of times to get to one solution which satisfies 0 < a <= b <= c <= d <= (1/2)abc. The idea is to use (1) to fix the signs, (2) to order the four variables, and then (3) to make d smaller than (1/2)abc. But then that might change the order, so you might have to repeat (2) and (3) until you get the above inequalities, but since d keeps getting smaller at each step, and there are only finitely many small integers, this process will eventually stop. Next, you should prove that there is exactly ONE integer solution which satisfies 0 < a <= b <= c <= d <= (1/2)abc. Proving this requires a lot of inequalities to prove that a, b, and c can't be very large, and then you try the small cases and see which ones work. And then you're done! Do you see how proving the above things would prove that all solutions come from (2, 2, 2, 2) by using the transformations I described? 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: 06/12/2007 at 03:54:34 From: Shamik Subject: Thank you (Number Theory: (a^2 + b^2 + c^2 + d^2) = a*b*c*d) Docter Vogler, Thank you very much. It was a new learning for me on how to attempt these kind of problems. High regards, Shamik
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