Factorise 51985 - 1 as a product of three integers, each greater than 5100.
Solution
Answer: (5397 - 1)(5794 - 5596+ 3·5397 - 5199 + 1)(5794 + 5596 + 3·5397 + 5199 + 1).
Solution by Demetres Christofides
Put A = 5397. Then 51985 - 1 = (A - 1)(A4 + A3 + A2 + A + 1). Clearly (A - 1) is one of the desired factors.
We have (A2 + 3A + 1)2 = A4 + 6A3 + 11A2 + 6A + 1, so A4 + A3 + A2 + A + 1 = (A2 + 3A + 1)2 - 5A(A + 1)2. But 5A = B2, where B = 5199. So A4 + A3 + A2 + A + 1 = (A2 + 3A + 1 + BA + B)(A2 + 3A + 1 - BA - B).
It remains to check that A2 + 3A + 1 - BA - B > 5100. But obviously 5794 > 5596 and 5397 > 5199. So the factor is > 2·5397.
© John Scholes
jscholes@kalva.demon.co.uk
11 Sep 2002