Question

设a为质数，b,c为正整数，且满足9(2a+2b-c)的平方=509（4a+1022b-511c)且b-c=2，求a(b+c)的值

Answer 1

由9(2a+2b-c)² = 509(4a+1022b-511c), a, b, c都为整数.
比较两边质因子, 得509 | 2a+2b-c (509是质数).
可设2a+2b-c = 509k, 其中k为正整数(2a+2b-c > 0).
则4a+1022b-511c = 4a+4b-2c+509(2b-c) = 509(2k+2b-c).
代回得9k² = 2k+2b-c, 即档历返2b-c = 9k²-2k, 于是2a = 509k-(2b-c) = (511-9k)k.
由a为质数, 比较两边分行饥解式, 可能有以下几种情况:
k = 1, 511-9k = 2a, 得a = 251为质数.
k = 2, 511-9k = a, 得a = 493 = 17·29不为质数, 舍去.
k = a, 511-9k = 2, 无整数解, 舍去.
k = 2a, 511-9k = 1, 无整数解, 舍去.
于是只有烂裂k = 1, a = 251, 代回得2b-c = 9k²-2k = 7, 又b-c = 2, 得b = 5, c = 3.
a(b+c) = 251·8 = 2008.

Answer 2

是（2a+2b-c）²，还是[9（2a+2b-c）] ²

追问

（2a+2b-c）²