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如果你要計算的方法，看看這個可否幫倒你。

分子是 : 3^2+1, 4^2+1, 5^2+1, ... , (n+2)^2+1
分母是 : 2*1+1, 2*2+1, 2*3+1, ... , 2n+1

所以通項是 [(n+2)^2 + 1]/(2n + 1), 若它是一整數 p, 則
(n+2)^2 + 1 = 2np + p
==> n^2 + 2(2-p)n + (5-p) = 0
==> n = (p-2) + 根[(2-p)^2 - (5-p)]
因為 n 為一自然數，所以
(2-p)^2 - (5-p) = r^2 . . . (r 也是一自然數)
==> p^2 - 3p - 1 = r^2
==> 4p^2 - 12p - 4 = 4r^2
==> (2p - 3)^2 - 13 = (2r)^2
==> (2p - 3)^2 - (2r)^2 = 13
==> (2p - 3 - 2r)(2pm - 3 + 2r) = 1*13
所以
2p - 3 - 2r = 1 . . . (1) 及 
2p - 3 + 2r = 13 . . (2)
解 (1), (2), 得 p = 5, r = 3
p = 5 即 n = (5-2) + 根[(2-5)^2 - (5-5)] = 6
所以第六項是為一的整數項，項值是 5.

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