總是平方數
 

12 + 22 + 22 = 9 = 3 2

22 + 32 + 62 = 49 = 7 2

32 + 42 + 122 =169 = 13 2

 42 + 52 + 202 = 441 = 21
2

顯然,左式是 n2+(n+1)2+[n(n+1)]2,n是自然數。

n2+(n+1)2+[n(n+1)]2 = n2+n2+2n+1+n2(n2+2n+1)

= n4+2n3+3n2+2n+1 = (n4+2n3+n2)+(2n2+2n)+1

= n2(n+1)2+2n(n+1)+1 = [n(n+1)]2+2n(n+1)+12

= [n(n+1)+1]2 = ( n2+n+1)2

所以 n2+(n+1)2+[n(n+1)]2 總是平方數。


 


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