Arithmetic Integers (Posted on 2006-02-06)

	Submitted by K Sengupta
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Let us consider ay given arithmetic series A,A+B,------,A+nB where (A,B)=k. Then, substituting C=A/k and D = B/k, we obtain the arithmetic series C,C+D,------,C+nD where (C,D) = 1, son that, any given arithmetic series x' + p*y' with (x',y') =1 for p =0,1,......,n will generate the arithmetic series x+p*y where x = k*x' and y=k*y' and k > 1 for p=0,1,....,n. Accordingly, the required positive integers are x-y, x and x+y ( say), where we can assume without any loss of generality that (x,y) =1 Since, (x-y) must be positive, it follows that x > y. Hence, (x-y)^19 + x^19 = (x+y)^19 gives, x^19= 0(Mod y) Since, (x,y) =1 by assumption, it follows that y=1. Accordingly; (x-1)19 + x^19 = (x+1)^19 or, 2 = 0 (Mod x) Since, x must be greater than 1, it follows that x=2, and we observe that x=2 yields 1 + 2^19 = 3^19 Now, we know that, 1 + 2^P < 3^P for all P greater than 2. Acordingly, 1 + 2^19 = 3^19 is a contradiction. Consequently, it is not feasible to determine three positive integers in Arithmetic Progression satisfying conditions of the problem. ----------Q E D ----------- For a general method for solving the problem under reference, refer to the mehodology provided below. GENERALISED SOLUTION: Suppose, we are given three positive integers x-y, x and x+y ( with x > y ) in Arithmetic Progression. We shall prove that no integral solution exists for the equation given below: (x-y)^P + x^P = (x +y)^P, where P is any positive integer greater than 2. PROOF: Without any loss of generality, we can assume that (x,y) = 1. Then, if possible, let: (x-y)^P + x^P = (x +y)^P,where P > 2 -------(#) or, x^P = 0 ( Mod y) or, y divides x^P which in conjunction with (x,y)=1 gives y =1. If x =1, then the smallest of the three numbers is x-y =1-1=0, which is a contradiction. For x=2, it follows that : 1 + 2^P = 3^P, for all integral P > 2 But we know that 1 + 2^P < 3^P for all integral P greater than 2. This is a contradiction. Hence, (x-1)^P + x^P = (x+1)^P, for P > 2 and x > 2 or, (-1)^P = 1 ( Mod x), so that P must be even and, since P > 2, it follows that P is grreater than or equal to 4. Accordingly; x^P = (x+1)^P - (x-1)^P for, x > 2 or, x^P = 2(P*(x^(P-1)) + P(P-1)(P-2)/6 *(x^(P-3)) + ------------------ + P*x) ( since P is even)--------------(##) Consequently, x must be even and P is an even number greater than or equal to 4; so that, 2*(x^3) divides x^4 and hence, 2*(x^3) divides x^P for all P greater than or equal to 4. Hence, from (##): 0 = 2*P*x ( Mod 2*(x^3) ) or, 2Px = 0 (Mod 2*(x^3)), so that 2*(x^3) divides 2Px, giving that P is greater than or equal to x^2. Now, (x-1)^P = (x+1)^P - x^P or,(x-1)^P > P*(x^(P-1)) > = x^ (P+1), which is not feasible and accordingly, this is a contradiction. Consequently, no positive integral solution exists for equation (#). Substituting P = 19, we obtain our desired result.

	Subject	Author	Date
	re: Proof, I think	Mindrod	2006-02-07 22:54:54
	Proof, I think	goFish	2006-02-06 16:01:36