One Good Factorization Deserves Another: 11029
Tim Keller, Doyle Henderson · American Mathematical Monthly · 2005
Solution by Doyle Henderson, University of Nebraska, Omaha, NE. The answers are 3, 3, and 2, respectively. When (a, b, c, d) = (11, 9, 6, 1), we have ab + cd = 3 • 5 • 7, ac+bd= (9 + 1+61)(9 + 1 1 + 6) = 352, and ad+bc= 5 13, sothese values are achievable, simultaneously. Let a = b + d + a c and / = b + d a + c. We first show that ad + bc is not prime. Since at = a(b + d + a a) + bd = aa + (a + b)(a + d), a I (a + b)(a + d). As a ]]> a + b, a shares a prime factor p1 with a + b. Since c d = (a + b) a, pl (c d). Now ad + bc = d(a + b) + b(c d), so pli (ad + bc). But pi -6 ad + bc because pi a while ad + bc = d(a + b) + b(c d) = da + (d + b)(c d) ]]> a. To complete the proof it suffices to obtain primes p2 and p3, not necessarily distinct, such that P2P3 properly divides both ab + cd and ac + bd. Now cd 2a. In tandem with the earlier result a I (a + b)(c + d), this forces a to share a prime factor P2 with a + d, and since b c = a (a + d), p2 I (b c) as well. From ab + cd = b(a + d) d(b c), we conclude that p2 I (ab + cd). Next, we note that f8 I (c + b)(c + d) because a3f = ac + bd = (b + d + c fl)c + bd = -fic + (b + c)(c + d). We claim that f/ and (b + c) have a common prime factor P3. Otherwise, /f I (c + d), and then f! I (a b) because a b = c + d ft. Let kl =