If r=125 ( rr4 + r2 + 1 ) = pq, where p and q are positive integers such that (p, q) = 1, then p + q is equal to .
Answer
976
Solution
r4 + r2 + 1 = (r2 + 1)2 - r2 = (r2 - r + 1)(r2 + r + 1)
rr4 + r2 + 1 = 12 [ 1r2 - r + 1 - 1r2 + r + 1 ]
(r + 1)2 - (r + 1) + 1 = r2 + r + 1, the sum telescopes:
r=125 = 12 [1 - 1252 + 25 + 1 ] = 12 [1 - 1651 ]
= 6501302 = 325651
325 = 52 13, 651 = 3 7 31 (325, 651) = 1
p + q = 325 + 651 = 976
Answer: 976