A Bivariate Divisibility Problem (Posted on 2006-06-08)

	Submitted by K Sengupta
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(p,q) = (55,1), (529,1), (43,2) and (23*(s^2), 23*s), where s is a positive integer, constitutes all possible solutions to the given problem. EXPLANATION: If p < q, then q > = p+1, giving: p*(q^2) + q + 23 > p*(q^2) + q >= (p+1)(pq+1) = (p^2)*q + p + pq > = (p^2)*q + p + q. Accordingly, no solution exists whenever p is less than q. If possible, let us assume that p >= q. Let s = ((p^2)*q + p + q) / (p*(q^2) + q + 23)-------(#). Accordingly, we obtain: (p/q + 1/q)*(p*(q^2) + q + 23) = (p^2)*q + p + pq + 23*p/q + 23/q + 1 > (p^2)*q + p + q So, s < p/q + 1/q. Now, if q >=5, then (q - 23/q) > 0, and hence: (p/q - 1/q)*(p*(q^2) + q + 23) = (p^2)*q + p - p(q - 23/q) - 1 - 23/q < (p^2)*q + p < (p^2)*q + p + q. Hence, either q =1,2,3,4 ; or, s > p/q - 1/q If, p/q - 1/q < s < p/q + 1/q; then, p-1 < qs < p+1. Hence, p=q*s. Accordingly, from (#), we obtain: (p,q)=(23*(s^2), 23s) We now proceed to consider the cases q =1,2,3,4 When, q = 1, then (p+24) divides (p^2)+ p + 1 Now, p(p+24) - ((p^2)+ p + 1) = 23*p -1, also, 23(p+24) - (23*p - 1)= 553 The only factors of 553 greater than 24 are 79 and 553, so that : p+ 24 = 79, 553 giving p = 55, 529; so that, (p,q) = (55,1), (529,1) When, q = 2, then (4p+25) divides 2*(p^2)+ p + 2 Now, p(4*p + 25) - 2(2*(p^2)+ p + 2) = 23*p - 4, also, 23(4*p+25) - 4(23*p - 4)= 591 The only factors of 591 greater than 25 are 197 and 591, so that: 4p+ 25 =197, 591 giving p = 43, 141.5; Hence, (p,q)= (43,2) for q=2 When, q = 3, then (9p+26) divides 3*(p^2)+ p + 3 Now, p(9*p + 26) - 3(3*(p^2)+ p + 3) = 23*p -9, also, 23(9*p+26) - 9(23*p - 9)= 679 The only factors of 679 greater than 26 are 97 and 679, so that : 9*p+ 26 =97, 679 giving p =71/9, 653/9; neither of which is a whole number and this is accordingly, a contradiction. Hence, no integral solution for p exists for q=3 When, q = 4, then (16*p + 27) divides 4*(p^2)+ p + 4 Now, p(16*p + 27) - 4(4*(p^2)+ p + 4) = 23*p - 16 also, 23(16*p + 27) - 16(23*p - 16)= 877 The only factor of 877 greater than 27 is 877, so that 16*p+ 27 =877 giving p =53.125, which is not a whole number and this is accordingly, a contradiction. Hence, no integral solution for p exists for q=4 Consequently, (p,q) = (55,1), (529,1), (43,2) and (23*(s^2), 23*s), where s is a positive integer, constitutes all possible solutions to the given problem. ALTERNATE SOLUTION: We have p*q^2 + q + 23 divides p(p*q^2 + q + 23) - q(p^2*q + p + q) = 23p – q^2 . If 23p =q^2, then q must be a multiple of 23, so q = 23k for some k. Then a = 23*(k^2), and it is easy to check that this is a solution. We cannot have 23*p < q^2 for then 0 < q^2 – 23p < p*q^2 < p*q^2 + q + 23. If 23p > q, then we must have 23p - q < = p*q^2 + q + 23 > p*q^2, so 23 > q^2, so q = 1,2,3,4. Continuing as before in the previous method, we observe that :(p,q) = (55,1), (529,1), (43,2) and (23*(s^2), 23*s), where s is a positive integer, constitutes all possible solutions to the given problem.

	Subject	Author	Date
	No Subject	K Sengupta	2006-06-08 23:26:37