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(接上楼)
(3)2ac^2-at^2+2a-2=t^2,t^2=2*(ac^2+a-1)/(a+1);
令a=3,t^2=2*(3c^2+3-1)/(3+1)=(3c^2+2)/2
a c t
3 4 5
3 40 49
3 396 485
或令a=-3,t^2=2*(-3c^2-3-1)/(-3+1)=3c^2+4
a c t
-3 2 4
-3 8 14
-3 30 52
-3 112 194
-3 418 724
2ac^2-at^2+2a-2=t^2,t^2=2*(ac^2+a-1)/(a+1);
令a=3,t^2=2*(3c^2+3-1)/(3+1)=(3c^2+2)/2
c=±((5-2sqrt(6))^n-(5+2sqrt(6))^n)/sqrt(6),
t=±1/2((5-2sqrt(6))^n+(5+2sqrt(6))^n),n∈Z,n>=0
令a=-3,t^2=2*(-3c^2-3-1)/(-3+1)=3c^2+4
c=±((2-sqrt(3))^n-(2+sqrt(3))^n)/sqrt(3),
t=-(2-sqrt(3))^n-(2+sqrt(3))^n,n∈Z,n>=0
c=±((2-sqrt(3))^n-(2+sqrt(3))^n)/sqrt(3),
t=(2-sqrt(3))^n+(2+sqrt(3))^n,n∈Z,n>=0
仅a=±3,就要无穷多组整数解,其中哪能都是素数?
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