由 danny 於 星期一 七月 30, 2012 11:20 pm
65=13*5,and let f(x)=5x^13+13x^5+9ax
obviously x=65k be a solution,k belongs to Z
by fermat's little theorem x^(p-1)=1 mod p , p is prime
hence, 5|f(x) => 5|(3x^5+9ax) => 5|x(3x^4+4a) => 4a=2 mod 5 ...(a)
similarly, 13|f(x) => 4a=5 mod 13 ...(b)
(a),(b) => 4a=-8 mod 65
a=65s-2 , s belongs to Z
so a=63