已知p,q均为质数,且满足7p²+3q=已知p,q均为质数,且满足7p²+3q=79 p= q=
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![已知p,q均为质数,且满足7p²+3q=已知p,q均为质数,且满足7p²+3q=79 p= q=](/uploads/image/z/3812263-7-3.jpg?t=%E5%B7%B2%E7%9F%A5p%2Cq%E5%9D%87%E4%B8%BA%E8%B4%A8%E6%95%B0%2C%E4%B8%94%E6%BB%A1%E8%B6%B37p%26%23178%3B%2B3q%3D%E5%B7%B2%E7%9F%A5p%2Cq%E5%9D%87%E4%B8%BA%E8%B4%A8%E6%95%B0%EF%BC%8C%E4%B8%94%E6%BB%A1%E8%B6%B37p%26%23178%3B%2B3q%3D79+p%3D+q%3D)
已知p,q均为质数,且满足7p²+3q=已知p,q均为质数,且满足7p²+3q=79 p= q=
已知p,q均为质数,且满足7p²+3q=
已知p,q均为质数,且满足7p²+3q=79 p= q=
已知p,q均为质数,且满足7p²+3q=已知p,q均为质数,且满足7p²+3q=79 p= q=
79为奇数,所以左边应该是奇数加偶数,p或q中有一为偶数(2)
若p=2,不成立
所以p=2,q=17
P=2 q=17
题目不全吧。