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

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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

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2b-c = 9k²-2k,于是2a = 509k-(2b-c) = (511-9k)k.
由a为质数,比较两边分解式,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.

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