回复：(n^2+1)^2-(n^2+2)*n^2=1,(n^2-1)^2-(n^2-2)*n^2=1.If for any co

I agree with you on those two cases about value of P.

By checking there is no y such that |y^2 -9781| == 2, we can only be sure that

Pi1^e1*...Pim^em > 1? but we still need to prove:

(9781Pi1^c1*...Pim^cm-Pi1^d1*...Pim^dm) > 1 when Pi1^e1*...Pim^em = 2.

Notice that generally this might not be true, e.g. 9781 - 2^2*5*3*163 = 1. or 9781*2 - 31*631 =1. The good thing is in this problem, t1 + s1 = 2a1, ... tm + sm = 2am. How can you prove this in tight math? Thanks.

所有跟帖： 

• Your Ex,5should be 5^2m,3 ...3^2n...,We get |9781-Z^2|=1, -jinjing- ♀ (344 bytes) () 01/27/2011 postreply 09:35:02

• 回复：Your Ex,5should be 5^2m,3 ...3^2n...,We get |9781-Z^2|=1, -calligraphy- ♂ (425 bytes) () 01/27/2011 postreply 11:26:44

• 回复：回复：Your Ex,5should be 5^2m,3 ...3^2n...,We get |9781-Z^2|=1, -jinjing- ♀ (438 bytes) () 01/27/2011 postreply 17:19:24