证明对于任意正整数a,【a、4a+4、9a+6、(8a+4)(6a+5)(3a+1)】满足【任意两数之积加上1都是平方数】这个规律:
a(4a+4)+1 = 4a²+4a+1 = (2a+1)²
a(9a+6)+1 = 9a²+6a+1 = (3a+1)²
(4a+4)(9a+6)+1 = 36a²+60a+25 = (6a+5)²
a(8a+4)(6a+5)(3a+1)+1 = (8a²+4a)(18a²+21a+5)+1
= 144a⁴+240a³+124a²+20a+1 = (12a²+10a+1)²
(4a+4)(8a+4)(6a+5)(3a+1)+1 = (24a²+20a+4)(24a²+44a+20)+1
= 576a⁴+1536a³+1456a²+576a+81 = (24a²+32a+9)²
(9a+6)(8a+4)(6a+5)(3a+1)+1 = (72a²+84a+24)(18a²+21a+5)+1
= 1296a⁴+3024a³+2556a²+924a+121 = (36a²+42a+11)²
![](https://tb2.bdstatic.com/tb/editor/images/face/i_f42.png?t=20140803)